1 First-Order Differential Equations

1.1 Differential Equations and Mathematical Models

The laws of the universe are written in the language of mathematics. Algebra is sufficient to solve many static problems, but the most interesting natural phenomena involve change and are described by equations that relate changing quantities.

Because the derivative dx/dt=f(t) of the function f is the rate at which the quantity x=f(t) is changing with respect to the independent variable t, it is natural that equations involving derivatives are frequently used to describe the changing universe. An equation relating an unknown function and one or more of its derivatives is called a differential equation.

Example 1

The differential equation

dxdt=x2+t2

involves both the unknown function x(t) and its first derivative x(t)=dx/dt. The differential equation

d2ydx2+3dydx+7y=0

involves the unknown function y of the independent variable x and the first two derivatives y and y of y.

The study of differential equations has three principal goals:

  1. To discover the differential equation that describes a specified physical situation.

  2. To find—either exactly or approximately—the appropriate solution of that equation.

  3. To interpret the solution that is found.

In algebra, we typically seek the unknown numbers that satisfy an equation such as x3+7x211x+41=0. By contrast, in solving a differential equation, we are challenged to find the unknown functions y=y(x) for which an identity such as y(x)=2xy(x)—that is, the differential equation

dydx=2xy

—holds on some interval of real numbers. Ordinarily, we will want to find all solutions of the differential equation, if possible.

Example 2

If C is a constant and

(1)y(x)=Cex2,

then

dydx=C(2xex2)=(2x)(Cex2)=2xy.

Thus every function y(x) of the form in Eq. (1) satisfies—and thus is a solution of—the differential equation

(2)dydx=2xy

for all x. In particular, Eq. (1) defines an infinite family of different solutions of this differential equation, one for each choice of the arbitrary constant C. By the method of separation of variables (Section 1.4) it can be shown that every solution of the differential equation in (2) is of the form in Eq. (1).

Differential Equations and Mathematical Models

The following three examples illustrate the process of translating scientific laws and principles into differential equations. In each of these examples the independent variable is time t, but we will see numerous examples in which some quantity other than time is the independent variable.

Example 3

Rate of cooling Newton’s law of cooling may be stated in this way: The time rate of change (the rate of change with respect to time t) of the temperature T(t) of a body is proportional to the difference between T and the temperature A of the surrounding medium (Fig. 1.1.1). That is,

(3)dTdt=k(TA),

where k is a positive constant. Observe that if T>A, then dT/dt<0, so the temperature is a decreasing function of t and the body is cooling. But if T<A, then dT/dt>0, so that T is increasing.

Thus the physical law is translated into a differential equation. If we are given the values of k and A, we should be able to find an explicit formula for T(t), and then—with the aid of this formula—we can predict the future temperature of the body.

FIGURE 1.1.1.

Newton’s law of cooling, Eq. (3), describes the cooling of a hot rock in water.

Example 4

Draining tank Torricelli’s law implies that the time rate of change of the volume V of water in a draining tank (Fig. 1.1.2) is proportional to the square root of the depth y of water in the tank:

(4)dVdt=ky,

where k is a constant. If the tank is a cylinder with vertical sides and cross-sectional area A, then V=Ay, so dV/dt=A·(dy/dt). In this case Eq. (4) takes the form

(5)dydt=hy,

where h=k/A is a constant.

FIGURE 1.1.2.

Torricelli’s law of draining, Eq. (4), describes the draining of a water tank.

Example 5

Population growth The time rate of change of a population P(t) with constant birth and death rates is, in many simple cases, proportional to the size of the population. That is,

(6)dPdt=kP,

where k is the constant of proportionality.

Let us discuss Example 5 further. Note first that each function of the form

(7)P(t)=Cekt

is a solution of the differential equation

dPdt=kP

in (6). We verify this assertion as follows:

P(t)=Ckekt=k(Cekt)=kP(t)

for all real numbers t. Because substitution of each function of the form given in (7) into Eq. (6) produces an identity, all such functions are solutions of Eq. (6).

Thus, even if the value of the constant k is known, the differential equation dP/dt=kP has infinitely many different solutions of the form P(t)=Cekt, one for each choice of the “arbitrary” constant C. This is typical of differential equations. It is also fortunate, because it may allow us to use additional information to select from among all these solutions a particular one that fits the situation under study.

Example 6

Population growth Suppose that P(t)=Cekt is the population of a colony of bacteria at time t, that the population at time t=0 (hours, h) was 1000, and that the population doubled after 1 h. This additional information about P(t) yields the following equations:

1000=P(0)=Ce0=C,2000=P(1)=Cek.

It follows that C=1000 and that ek=2, so k=ln 20.693147. With this value of k the differential equation in (6) is

dPdt=(ln 2)P(0.693147)P.

Substitution of k=ln 2 and C=1000 in Eq. (7) yields the particular solution

P(t)=1000e(ln 2)t=1000(eln 2)t=10002t(because eln 2=2)

that satisfies the given conditions. We can use this particular solution to predict future populations of the bacteria colony. For instance, the predicted number of bacteria in the population after one and a half hours (when t=1.5) is

P(1.5)=100023/22828.

The condition P(0)=1000 in Example 6 is called an initial condition because we frequently write differential equations for which t=0 is the “starting time.” Figure 1.1.3 shows several different graphs of the form P(t)=Cekt with k=ln 2. The graphs of all the infinitely many solutions of dP/dt=kP in fact fill the entire two-dimensional plane, and no two intersect. Moreover, the selection of any one point P0 on the P-axis amounts to a determination of P(0). Because exactly one solution passes through each such point, we see in this case that an initial condition P(0)=P0 determines a unique solution agreeing with the given data.

FIGURE 1.1.3.

Graphs of P(t)=Cekt with k=ln 2.

Mathematical Models

Our brief discussion of population growth in Examples 5 and 6 illustrates the crucial process of mathematical modeling (Fig. 1.1.4), which involves the following:

  1. The formulation of a real-world problem in mathematical terms; that is, the construction of a mathematical model.

  2. The analysis or solution of the resulting mathematical problem.

  3. The interpretation of the mathematical results in the context of the original real-world situation—for example, answering the question originally posed.

FIGURE 1.1.4.

The process of mathematical modeling.

In the population example, the real-world problem is that of determining the population at some future time. A mathematical model consists of a list of variables (P and t) that describe the given situation, together with one or more equations relating these variables (dP/dt=kP, P(0)=P0) that are known or are assumed to hold. The mathematical analysis consists of solving these equations (here, for P as a function of t). Finally, we apply these mathematical results to attempt to answer the original real-world question.

As an example of this process, think of first formulating the mathematical model consisting of the equations dP/dt=kP, P(0)=1000, describing the bacteria population of Example 6. Then our mathematical analysis there consisted of solving for the solution function P(t)=1000e(ln 2)t=1000·2t as our mathematical result. For an interpretation in terms of our real-world situation—the actual bacteria population—we substituted t=1.5 to obtain the predicted population of P(1.5)2828 bacteria after 1.5 hours. If, for instance, the bacteria population is growing under ideal conditions of unlimited space and food supply, our prediction may be quite accurate, in which case we conclude that the mathematical model is adequate for studying this particular population.

On the other hand, it may turn out that no solution of the selected differential equation accurately fits the actual population we’re studying. For instance, for no choice of the constants C and k does the solution P(t)=Cekt in Eq. (7) accurately describe the actual growth of the human population of the world over the past few centuries. We must conclude that the differential equation dP/dt=kP is inadequate for modeling the world population—which in recent decades has “leveled off” as compared with the steeply climbing graphs in the upper half (P>0) of Fig. 1.1.3. With sufficient insight, we might formulate a new mathematical model including a perhaps more complicated differential equation, one that takes into account such factors as a limited food supply and the effect of increased population on birth and death rates. With the formulation of this new mathematical model, we may attempt to traverse once again the diagram of Fig. 1.1.4 in a counterclockwise manner. If we can solve the new differential equation, we get new solution functions to compare with the real-world population. Indeed, a successful population analysis may require refining the mathematical model still further as it is repeatedly measured against real-world experience.

But in Example 6 we simply ignored any complicating factors that might affect our bacteria population. This made the mathematical analysis quite simple, perhaps unrealistically so. A satisfactory mathematical model is subject to two contradictory requirements: It must be sufficiently detailed to represent the real-world situation with relative accuracy, yet it must be sufficiently simple to make the mathematical analysis practical. If the model is so detailed that it fully represents the physical situation, then the mathematical analysis may be too difficult to carry out. If the model is too simple, the results may be so inaccurate as to be useless. Thus there is an inevitable tradeoff between what is physically realistic and what is mathematically possible. The construction of a model that adequately bridges this gap between realism and feasibility is therefore the most crucial and delicate step in the process. Ways must be found to simplify the model mathematically without sacrificing essential features of the real-world situation.

Mathematical models are discussed throughout this book. The remainder of this introductory section is devoted to simple examples and to standard terminology used in discussing differential equations and their solutions.

Examples and Terminology

Example 7

If C is a constant and y(x)=1/(Cx), then

dydx=1(Cx)2=y2

if xC. Thus

(8)y(x)=1Cx

defines a solution of the differential equation

(9)dydx=y2

on any interval of real numbers not containing the point x=C. Actually, Eq. (8) defines a one-parameter family of solutions of dy/dx=y2, one for each value of the arbitrary constant or “parameter” C. With C=1 we get the particular solution

y(x)=11x

that satisfies the initial condition y(0)=1. As indicated in Fig. 1.1.5, this solution is continuous on the interval (,1) but has a vertical asymptote at x=1.

Example 8

Verify that the function y(x)=2x1/2x1/2 ln x satisfies the differential equation

(10)4x2y+y=0

for all x>0.

Solution

First we compute the derivatives

y(x)=12x1/2 ln xandy(x)=14x3/2 ln x12x3/2.

Then substitution into Eq. (10) yields

4x2y+y=4x2(14x3/2 ln x12x3/2)+2x1/2x1/2 ln  x=0

if x is positive, so the differential equation is satisfied for all x>0.

The fact that we can write a differential equation is not enough to guarantee that it has a solution. For example, it is clear that the differential equation

(11)(y)2+y2=1

has no (real-valued) solution, because the sum of nonnegative numbers cannot be negative. For a variation on this theme, note that the equation

(12)(y)2+y2=0

obviously has only the (real-valued) solution y(x)0. In our previous examples any differential equation having at least one solution indeed had infinitely many.

The order of a differential equation is the order of the highest derivative that appears in it. The differential equation of Example 8 is of second order, those in Examples 2 through 7 are first-order equations, and

y(4)+x2y(3)+x5y=sin x

is a fourth-order equation. The most general form of an nth-order differential equation with independent variable x and unknown function or dependent variable y=y(x) is

(13)F(x, y, y, y, , y(n))=0,

where F is a specific real-valued function of n+2 variables.

Our use of the word solution has been until now somewhat informal. To be precise, we say that the continuous function u=u(x) is a solution of the differential equation in (13) on the interval I provided that the derivatives u, u, , u(n) exist on I and

F(x, u, u, u, , u(n))=0

for all x in I. For the sake of brevity, we may say that u=u(x) satisfies the differential equation in (13) on I.

Remark

Recall from elementary calculus that a differentiable function on an open interval is necessarily continuous there. This is why only a continuous function can qualify as a (differentiable) solution of a differential equation on an interval.

FIGURE 1.1.5.

The solution of y=y2 defined by y(x)=1/(1x).

Example 7

Continued Figure 1.1.5 shows the two “connected” branches of the graph y=1/(1x). The left-hand branch is the graph of a (continuous) solution of the differential equation y=y2 that is defined on the interval (, 1). The right-hand branch is the graph of a different solution of the differential equation that is defined (and continuous) on the different interval (1, ). So the single formula y(x)=1/(1x) actually defines two different solutions (with different domains of definition) of the same differential equation y=y2.

Example 9

If A and B are constants and

(14)y(x)=A cos 3x+B sin 3x.

then two successive differentiations yield

y(x)=3A sin 3x+3B cos 3x,y(x)=9A cos 3x9B sin 3x=9y(x)

for all x. Consequently, Eq. (14) defines what it is natural to call a two-parameter family of solutions of the second-order differential equation

(15)y+9y=0

on the whole real number line. Figure 1.1.6 shows the graphs of several such solutions.

Although the differential equations in (11) and (12) are exceptions to the general rule, we will see that an nth-order differential equation ordinarily has an n-parameter family of solutions—one involving n different arbitrary constants or parameters.

FIGURE 1.1.6.

The three solutions y1(x)=3 cos 3x, y2(x)=2 sin 3x, and y3(x)=3 cos 3x+2 sin 3x of the differential equation y+9y=0.

In both Eqs. (11) and (12), the appearance of y as an implicitly defined function causes complications. For this reason, we will ordinarily assume that any differential equation under study can be solved explicitly for the highest derivative that appears; that is, that the equation can be written in the so-called normal form

(16)y(n)=G(x, y, y, y, , y(n1)),

where G is a real-valued function of n+1 variables. In addition, we will always seek only real-valued solutions unless we warn the reader otherwise.

All the differential equations we have mentioned so far are ordinary differential equations, meaning that the unknown function (dependent variable) depends on only a single independent variable. If the dependent variable is a function of two or more independent variables, then partial derivatives are likely to be involved; if they are, the equation is called a partial differential equation. For example, the temperature u=u(x,t) of a long thin uniform rod at the point x at time t satisfies (under appropriate simple conditions) the partial differential equation

ut=k2ux2,

where k is a constant (called the thermal diffusivity of the rod). In Chapters 1 through 8 we will be concerned only with ordinary differential equations and will refer to them simply as differential equations.

In this chapter we concentrate on first-order differential equations of the form

(17)dydx=f(x, y).

We also will sample the wide range of applications of such equations. A typical mathematical model of an applied situation will be an initial value problem, consisting of a differential equation of the form in (17) together with an initial condition y(x0)=y0. Note that we call y(x0)=y0 an initial condition whether or not x0=0. To solve the initial value problem

(18)dydx=f(x, y),y(x0)=y0

means to find a differentiable function y=y(x) that satisfies both conditions in Eq. (18) on some interval containing x0.

Example 10

Given the solution y(x)=1/(Cx) of the differential equation dy/dx=y2 discussed in Example 7, solve the initial value problem

dydx=y2,   y(1)=2.

Solution

We need only find a value of C so that the solution y(x)=1/(Cx) satisfies the initial condition y(1)=2. Substitution of the values x=1 and y=2 in the given solution yields

2=y(1)=1C1,

so 2C2=1, and hence C=32. With this value of C we obtain the desired solution

y(x)=132x=232x.

Figure 1.1.7 shows the two branches of the graph y=2/(32x). The left-hand branch is the graph on (,32) of the solution of the given initial value problem y=y2, y(1)=2. The right-hand branch passes through the point (2,2) and is therefore the graph on (32,) of the solution of the different initial value problem y=y2, y(2)=2.

The central question of greatest immediate interest to us is this: If we are given a differential equation known to have a solution satisfying a given initial condition, how do we actually find or compute that solution? And, once found, what can we do with it? We will see that a relatively few simple techniques—separation of variables (Section 1.4), solution of linear equations (Section 1.5), elementary substitution methods (Section 1.6)—are enough to enable us to solve a variety of first-order equations having impressive applications.

FIGURE 1.1.7.

The solutions of y=y2 defined by y(x)=2/(32x).

1.1 Problems

In Problems 1 through 12, verify by substitution that each given function is a solution of the given differential equation. Throughout these problems, primes denote derivatives with respect to x.

  1. y=3x2; y=x3+7

  2. y+2y=0; y=3e2x

  3. y+4y=0; y1=cos 2x, y2=sin 2x

  4. y=9y; y1=e3x, y2=e3x

  5. y=y+2ex; y=exex

  6. y+4y+4y=0; y1=e2x, y2=xe2x

  7. y2y+2y=0; y1=ex cos x, y2=ex sin x

  8. y+y=3 cos 2x, y1=cos xcos 2x, y2=sin xcos 2x

  9. y+2xy2=0; y=11+x2

  10. x2y+xyy=ln x; y1=xln x, y2=1xln x

  11. x2y+5xy+4y=0; y1=1x2, y2=ln xx2

  12. x2yxy+2y=0; y1=xcos(ln x), y2=xsin(ln x)

In Problems 13 through 16, substitute y=erx into the given differential equation to determine all values of the constant r for which y=erx is a solution of the equation.

  1. 3y=2y

  2. 4y=y

  3. y+y2y=0

  4. 3y+3y4y=0

In Problems 17 through 26, first verify that y(x) satisfies the given differential equation. Then determine a value of the constant C so that y(x) satisfies the given initial condition. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation, and highlight the one that satisfies the given initial condition.

  1. y+y=0; y(x)=Cex, y(0)=2

  2. y=2y; y(x)=Ce2x, y(0)=3

  3. y=y+1; y(x)=Cex1, y(0)=5

  4. y=xy; y(x)=Cex+x1, y(0)=10

  5. y+3x2y=0; y(x)=Cex3, y(0)=7

  6. eyy=1; y(x)=ln (x+C), y(0)=0

  7. xdydx+3y=2x5; y(x)=14x5+Cx3, y(2)=1

  8. xy3y=x3; y(x)=x3(C+ln x), y(1)=17

  9. y=3x2(y2+1); y(x)=tan(x3+C), y(0)=1

  10. y+ytanx=cos x; y(x)=(x+C) cos x, y(π)=0

In Problems 27 through 31, a function y=g(x) is described by some geometric property of its graph. Write a differential equation of the form dy/dx=f(x,y) having the function g as its solution (or as one of its solutions).

  1. The slope of the graph of g at the point (x, y) is the sum of x and y.

  2. The line tangent to the graph of g at the point (x, y) intersects the x-axis at the point (x/2,0).

  3. Every straight line normal to the graph of g passes through the point (0, 1). Can you guess what the graph of such a function g might look like?

  4. The graph of g is normal to every curve of the form y=x2+k (k is a constant) where they meet.

  5. The line tangent to the graph of g at (x, y) passes through the point (y,x).

Differential Equations as Models

In Problems 32 through 36, write—in the manner of Eqs. (3) through (6) of this section—a differential equation that is a mathematical model of the situation described.

  1. The time rate of change of a population P is proportional to the square root of P.

  2. The time rate of change of the velocity v of a coasting motorboat is proportional to the square of v.

  3. The acceleration dv/dt of a Lamborghini is proportional to the difference between 250 km/h and the velocity of the car.

  4. In a city having a fixed population of P persons, the time rate of change of the number N of those persons who have heard a certain rumor is proportional to the number of those who have not yet heard the rumor.

  5. In a city with a fixed population of P persons, the time rate of change of the number N of those persons infected with a certain contagious disease is proportional to the product of the number who have the disease and the number who do not.

In Problems 37 through 42, determine by inspection at least one solution of the given differential equation. That is, use your knowledge of derivatives to make an intelligent guess. Then test your hypothesis.

  1. y=0

  2. y=y

  3. xy+y=3x2

  4. (y)2+y2=1

  5. y+y=ex

  6. y+y=0

Problems 43 through 46 concern the differential equation

dxdt=kx2,

where k is a constant.

    1. If k is a constant, show that a general (one-parameter) solution of the differential equation is given by x(t)=1/(Ckt), where C is an arbitrary constant.

    2. Determine by inspection a solution of the initial value problem x=kx2, x(0)=0.

    1. Assume that k is positive, and then sketch graphs of solutions of x=kx2 with several typical positive values of x(0).

    2. How would these solutions differ if the constant k were negative?

  1. Suppose a population P of rodents satisfies the differential equation dP/dt=kP2. Initially, there are P(0)=2 rodents, and their number is increasing at the rate of dP/dt=1 rodent per month when there are P=10 rodents. Based on the result of Problem 43, how long will it take for this population to grow to a hundred rodents? To a thousand? What’s happening here?

    FIGURE 1.1.8.

    Graphs of solutions of the equation dy/dx=y2.

  2. Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt=kv2. The initial speed of the motorboat is v(0)=10 meters per second (m/s), and v is decreasing at the rate of 1 m/s2 when v=5 m/s. Based on the result of Problem 43, long does it take for the velocity of the boat to decrease to 1 m/s? To 110 m/s? When does the boat come to a stop?

  3. In Example 7 we saw that y(x)=1/(Cx) defines a one-parameter family of solutions of the differential equation dy/dx=y2.

    1. Determine a value of C so that y(10)=10.

    2. Is there a value of C such that y(0)=0? Can you nevertheless find by inspection a solution of dy/dx=y2 such that y(0)=0?

    3. Figure 1.1.8 shows typical graphs of solutions of the form y(x)=1/(Cx). Does it appear that these solution curves fill the entire xy-plane? Can you conclude that, given any point (a, b) in the plane, the differential equation dy/dx=y2 has exactly one solution y(x) satisfying the condition y(a)=b?

    1. Show that y(x)=Cx4 defines a one-parameter family of differentiable solutions of the differential equation xy=4y (Fig. 1.1.9).

    2. Show that

      y(x)={x4if x<0,x4if x0

      defines a differentiable solution of xy=4y for all x, but is not of the form y(x)=Cx4.

    3. Given any two real numbers a and b, explain why—in contrast to the situation in part (c) of Problem 47—there exist infinitely many differentiable solutions of xy=4y that all satisfy the condition y(a)=b.

    FIGURE 1.1.9.

    The graph y=Cx4 for various values of C.

1.2 Integrals as General and Particular Solutions

The first-order equation dy/dx=f(x, y) takes an especially simple form if the right-hand-side function f does not actually involve the dependent variable y, so

(1)dydx=f(x).

In this special case we need only integrate both sides of Eq. (1) to obtain

(2)y(x)=f(x)dx+C.

This is a general solution of Eq. (1), meaning that it involves an arbitrary constant C, and for every choice of C it is a solution of the differential equation in (1). If G(x) is a particular antiderivative of f—that is, if G(x)f(x)—then

(3)y(x)=G(x)+C.

FIGURE 1.2.1.

Graphs of y=14x2+C for various values of C.

FIGURE 1.2.2.

Graphs of y=sin x+C for various values of C.

The graphs of any two such solutions y1(x)=G(x)+C1 and y2(x)=G(x)+C2 on the same interval I are “parallel” in the sense illustrated by Figs. 1.2.1 and 1.2.2. There we see that the constant C is geometrically the vertical distance between the two curves y(x)=G(x) and y(x)=G(x)+C.

To satisfy an initial condition y(x0)=y0, we need only substitute x=x0 and y=y0 into Eq. (3) to obtain y0=G(x0)+C, so that C=y0G(x0). With this choice of C, we obtain the particular solution of Eq. (1) satisfying the initial value problem

dydx=f(x),y(x0)=y0.

We will see that this is the typical pattern for solutions of first-order differential equations. Ordinarily, we will first find a general solution involving an arbitrary constant C. We can then attempt to obtain, by appropriate choice of C, a particular solution satisfying a given initial condition y(x0)=y0.

Remark

As the term is used in the previous paragraph, a general solution of a first-order differential equation is simply a one-parameter family of solutions. A natural question is whether a given general solution contains every particular solution of the differential equation. When this is known to be true, we call it the general solution of the differential equation. For example, because any two antiderivatives of the same function f(x) can differ only by a constant, it follows that every solution of Eq. (1) is of the form in (2). Thus Eq. (2) serves to define the general solution of (1).

Example 1

Solve the initial value problem

dydx=2x+3,y(1)=2.

Solution

Integration of both sides of the differential equation as in Eq. (2) immediately yields the general solution

y(x)=(2x+3)dx=x2+3x+C.

Figure 1.2.3 shows the graph y=x2+3x+C for various values of C. The particular solution we seek corresponds to the curve that passes through the point (1, 2), thereby satisfying the initial condition

y(1)=(1)2+3(1)+C=2.

It follows that C=2, so the desired particular solution is

y(x)=x2+3x2.

Second-order equations. The observation that the special first-order equation dy/dx=f(x) is readily solvable (provided that an antiderivative of f can be found) extends to second-order differential equations of the special form

(4)d2ydx2=g(x),

FIGURE 1.2.3.

Solution curves for the differential equation in Example 1.

in which the function g on the right-hand side involves neither the dependent variable y nor its derivative dy/dx. We simply integrate once to obtain

dydx=y(x)dx=g(x)dx=G(x)+C1,

where G is an antiderivative of g and C1 is an arbitrary constant. Then another integration yields

y(x)=y(x) dx=[G(x)+C1] dx=G(x) dx+C1x+C2,

where C2 is a second arbitrary constant. In effect, the second-order differential equation in (4) is one that can be solved by solving successively the first-order equations

dvdx=g(x)anddydx=υ(x).

Velocity and Acceleration

Direct integration is sufficient to allow us to solve a number of important problems concerning the motion of a particle (or mass point) in terms of the forces acting on it. The motion of a particle along a straight line (the x-axis) is described by its position function

(5)x=f(t)

giving its x-coordinate at time t. The velocity of the particle is defined to be

(6)υ(t)=f(t);that is,υ=dxdt.

Its acceleration a(t) is a(t)=v(t)=x(t); in Leibniz notation,

(7)a=dυdt=d2xdt2.

Equation (6) is sometimes applied either in the indefinite integral form x(t)=v(t)dt or in the definite integral form

x(t)=x(t0)+t0tυ(s)ds,

which you should recognize as a statement of the fundamental theorem of calculus (precisely because dx/dt=v).

Newton’s second law of motion says that if a force F(t) acts on the particle and is directed along its line of motion, then

(8)ma(t)=F(t);that is,F=ma,

where m is the mass of the particle. If the force F is known, then the equation x(t)=F(t)/m can be integrated twice to find the position function x(t) in terms of two constants of integration. These two arbitrary constants are frequently determined by the initial position x0=x(0) and the initial velocity v0=v(0) of the particle.

Constant acceleration. For instance, suppose that the force F, and therefore the acceleration a=F/m, are constant. Then we begin with the equation

(9)dυdt=a(a is a constant)

and integrate both sides to obtain

υ(t)=a dt=at+C1.

We know that v=v0 when t=0, and substitution of this information into the preceding equation yields the fact that C1=v0. So

(10)v(t)=dxdt=at+υ0.

A second integration gives

x(t)=υ(t) dt=(at+υ0) dt=12at2+υ0t+C2,

and the substitution t=0, x=x0 gives C2=x0. Therefore,

(11)x(t)=12at2+υ0t+x0.

Thus, with Eq. (10) we can find the velocity, and with Eq. (11) the position, of the particle at any time t in terms of its constant acceleration a, its initial velocity v0, and its initial position x0.

Example 2

Lunar lander A lunar lander is falling freely toward the surface of the moon at a speed of 450 meters per second (m/s). Its retrorockets, when fired, provide a constant deceleration of 2.5 meters per second per second (m/s2) (the gravitational acceleration produced by the moon is assumed to be included in the given deceleration). At what height above the lunar surface should the retrorockets be activated to ensure a “soft touchdown” (v=0 at impact)?

Solution

We denote by x(t) the height of the lunar lander above the surface, as indicated in Fig. 1.2.4. We let t=0 denote the time at which the retrorockets should be fired. Then v0=450 (m/s, negative because the height x(t) is decreasing), and a=+2.5, because an upward thrust increases the velocity v (although it decreases the speed |v|). Then Eqs. (10) and (11) become

FIGURE 1.2.4.

The lunar lander of Example 2.

(12)υ(t)=2.5t450

and

(13)x(t)=1.25t2450t+x0,

where x0 is the height of the lander above the lunar surface at the time t=0 when the retrorockets should be activated.

From Eq. (12) we see that v=0 (soft touchdown) occurs when t=450/2.5=180 s(that is, 3 minutes); then substitution of t=180, x=0 into Eq. (13) yields

x0=0(1.25)(180)2+450(180)=40,500

meters—that is, x0=40.5 km2516 miles. Thus the retrorockets should be activated when the lunar lander is 40.5 kilometers above the surface of the moon, and it will touch down softly on the lunar surface after 3 minutes of decelerating descent.

Physical Units

Numerical work requires units for the measurement of physical quantities such as distance and time. We sometimes use ad hoc units—such as distance in miles or kilometers and time in hours—in special situations (such as in a problem involving an auto trip). However, the foot-pound-second (fps) and meter-kilogram-second (mks) unit systems are used more generally in scientific and engineering problems. In fact, fps units are commonly used only in the United States (and a few other countries), while mks units constitute the standard international system of scientific units.

fps units mks units
Force pound (lb) newton (N)
Mass slug kilogram (kg)
Distance foot (ft) meter (m)
Time second (s) second (s)
g 32 ft/s2 9.8 m/s2

The last line of this table gives values for the gravitational acceleration g at the surface of the earth. Although these approximate values will suffice for most examples and problems, more precise values are 9.7805 m/s2 and 32.088 ft/s2 (at sea level at the equator).

Both systems are compatible with Newton’s second law F=ma. Thus 1 N is (by definition) the force required to impart an acceleration of 1 m/s2 to a mass of 1 kg. Similarly, 1 slug is (by definition) the mass that experiences an acceleration of 1 ft/s2 under a force of 1 lb. (We will use mks units in all problems requiring mass units and thus will rarely need slugs to measure mass.)

Inches and centimeters (as well as miles and kilometers) also are commonly used in describing distances. For conversions between fps and mks units it helps to remember that

1 in.=2.54 cm (exactly)and1 lb4.448 N.

For instance,

1 ft=12 in.×2.54cmin.=30.48 cm,

and it follows that

1 mi=5280 ft×30.48cmft=160934.4 cm1.609 km.

Thus a posted U.S. speed limit of 50 mi/h means that—in international terms—the legal speed limit is about 50×1.60980.45 km/h.

Vertical Motion with Gravitational Acceleration

The weight W of a body is the force exerted on the body by gravity. Substitution of a=g and F=W in Newton’s second law F=ma gives

(14)W=mg

for the weight W of the mass m at the surface of the earth (where g32 ft/s29.8 m/s2). For instance, a mass of m=20 kg has a weight of W=(20 kg)(9.8 m/s2)=196 N. Similarly, a mass m weighing 100 pounds has mks weight

W=(100 lb)(4.448 N/lb)=444.8 N,

so its mass is

m=Wg=444.8 N9.8 m/s245.4 kg.

To discuss vertical motion it is natural to choose the y-axis as the coordinate system for position, frequently with y=0 corresponding to “ground level.” If we choose the upward direction as the positive direction, then the effect of gravity on a vertically moving body is to decrease its height and also to decrease its velocity v=dy/dt. Consequently, if we ignore air resistance, then the acceleration a=dv/dt of the body is given by

(15)dυdt=g.

This acceleration equation provides a starting point in many problems involving vertical motion. Successive integrations (as in Eqs. (10) and (11)) yield the velocity and height formulas

(16)υ(t)=gt+υ0

and

(17)y(t)=12gt2+υ0t+y0.

Here, y0 denotes the initial (t=0) height of the body and v0 its initial velocity.

Example 3

Projectile motion

  1. Suppose that a ball is thrown straight upward from the ground (y0=0) with initial velocity v0=96 (ft/s, so we use g=32 ft/s2 in fps units). Then it reaches its maximum height when its velocity (Eq. (16)) is zero,

    υ(t)=32t+96=0,

    and thus when t=3 s. Hence the maximum height that the ball attains is

    y(3)=123232+963+0=144 (ft)

    (with the aid of Eq. (17)).

  2. If an arrow is shot straight upward from the ground with initial velocity v0=49 (m/s, so we use g=9.8 m/s2 in mks units), then it returns to the ground when

    y(t)=12(9.8)t2+49t=(4.9)t(t+10)=0,

    and thus after 10 s in the air.

A Swimmer’s Problem

Figure 1.2.5 shows a northward-flowing river of width w=2a. The lines x=±a represent the banks of the river and the y-axis its center. Suppose that the velocity vR at which the water flows increases as one approaches the center of the river, and indeed is given in terms of distance x from the center by

(18)υR=υ0(1x2a2).

You can use Eq. (18) to verify that the water does flow the fastest at the center, where vR=v0, and that vR=0 at each riverbank.

FIGURE 1.2.5.

A swimmer’s problem (Example 4).

Suppose that a swimmer starts at the point (a,0) on the west bank and swims due east (relative to the water) with constant speed vS. As indicated in Fig. 1.2.5, his velocity vector (relative to the riverbed) has horizontal component vS and vertical component vR. Hence the swimmer’s direction angle α is given by

tan α=vRvS.

Because tan α=dy/dx, substitution using (18) gives the differential equation

(19)dydx=υ0υS(1x2a2)

for the swimmer’s trajectory y=y(x) as he crosses the river.

Example 4

River crossing Suppose that the river is 1 mile wide and that its midstream velocity is v0=9 mi/h. If the swimmer’s velocity is vS=3 mi/h, then Eq. (19) takes the form

dydx=3(14x2).

Integration yields

y(x)=(312x2)dx=3x4x3+C

for the swimmer’s trajectory. The initial condition y(12)=0 yields C=1, so

y(x)=3x4x3+1.

Then

y(12)=3(12)4(12)3+1=2.

so the swimmer drifts 2 miles downstream while he swims 1 mile across the river.

1.2 Problems

In Problems 1 through 10, find a function y=f(x) satisfying the given differential equation and the prescribed initial condition.

  1. dydx=2x+1; y(0)=3

  2. dydx=(x2)2; y(2)=1

  3. dydx=x; y(4)=0

  4. dydx=1x2; y(1)=5

  5. dydx=1x+2; y(2)=1

  6. dydx=xx2+9; y(4)=0

  7. dydx=10x2+1; y(0)=0

  8. dydx=cos 2x; y(0)=1

  9. dydx=11x2; y(0)=0

  10. dydx=xex; y(0)=1

In Problems 11 through 18, find the position function x(t) of a moving particle with the given acceleration a(t), initial position x0=x(0), and initial velocity v0=v(0).

  1. a(t)=50, v0=10, x0=20

  2. a(t)=20, v0=15, x0=5

  3. a(t)=3t, v0=5, x0=0

  4. a(t)=2t+1, v0=7, x0=4

  5. a(t)=4(t+3)2, v0=1, x0=1

  6. a(t)=1t+4, v0=1, x0=1

  7. a(t)=1(t+1)3, v0=0, x0=0

  8. a(t)=50 sin 5t, v0=10, x0=8

Velocity Given Graphically

In Problems 19 through 22, a particle starts at the origin and travels along the x-axis with the velocity function v(t) whose graph is shown in Figs. 1.2.6 through 1.2.9. Sketch the graph of the resulting position function x(t) for 0t10.

  1. FIGURE 1.2.6.

    Graph of the velocity function υ(t) of Problem 19.

  2. FIGURE 1.2.7.

    Graph of the velocity function υ(t) of Problem 20.

  3. FIGURE 1.2.8.

    Graph of the velocity function υ(t) of Problem 21.

  4. FIGURE 1.2.9.

    Graph of the velocity function υ(t) of Problem 22.

Problems 23 through 28 explore the motion of projectiles under constant acceleration or deceleration.

  1. What is the maximum height attained by the arrow of part (b) of Example 3?

  2. A ball is dropped from the top of a building 400 ft high. How long does it take to reach the ground? With what speed does the ball strike the ground?

  3. The brakes of a car are applied when it is moving at 100 km/h and provide a constant deceleration of 10 meters per second per second (m/s2). How far does the car travel before coming to a stop?

  4. A projectile is fired straight upward with an initial velocity of 100 m/s from the top of a building 20 m high and falls to the ground at the base of the building. Find

    1. its maximum height above the ground;

    2. when it passes the top of the building;

    3. its total time in the air.

  5. A ball is thrown straight downward from the top of a tall building. The initial speed of the ball is 10 m/s. It strikes the ground with a speed of 60 m/s. How tall is the building?

  6. A baseball is thrown straight downward with an initial speed of 40 ft/s from the top of the Washington Monument (555 ft high). How long does it take to reach the ground, and with what speed does the baseball strike the ground?

  7. Variable acceleration A diesel car gradually speeds up so that for the first 10 s its acceleration is given by

    dυdt=(0.12)t2+(0.6)t(ft/s2).

    If the car starts from rest (x0=0, v0=0), find the distance it has traveled at the end of the first 10 s and its velocity at that time.

Problems 30 through 32 explore the relation between the speed of an auto and the distance it skids when the brakes are applied.

  1. A car traveling at 60 mi/h (88 ft/s) skids 176 ft after its brakes are suddenly applied. Under the assumption that the braking system provides constant deceleration, what is that deceleration? For how long does the skid continue?

  2. The skid marks made by an automobile indicated that its brakes were fully applied for a distance of 75 m before it came to a stop. The car in question is known to have a constant deceleration of 20 m/s2 under these conditions. How fast—in km/h—was the car traveling when the brakes were first applied?

  3. Suppose that a car skids 15 m if it is moving at 50 km/h when the brakes are applied. Assuming that the car has the same constant deceleration, how far will it skid if it is moving at 100 km/h when the brakes are applied?

Problems 33 and 34 explore vertical motion on a planet with gravitational acceleration different than the Earth’s.

  1. On the planet Gzyx, a ball dropped from a height of 20 ft hits the ground in 2 s. If a ball is dropped from the top of a 200-ft-tall building on Gzyx, how long will it take to hit the ground? With what speed will it hit?

  2. A person can throw a ball straight upward from the surface of the earth to a maximum height of 144 ft. How high could this person throw the ball on the planet Gzyx of Problem 33?

  3. Velocity in terms of height A stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes the ground is v=2gh.

  4. Varying gravitational acceleration Suppose a woman has enough “spring” in her legs to jump (on earth) from the ground to a height of 2.25 feet. If she jumps straight upward with the same initial velocity on the moon—where the surface gravitational acceleration is (approximately) 5.3 ft/s2—how high above the surface will she rise?

  5. At noon a car starts from rest at point A and proceeds at constant acceleration along a straight road toward point B. If the car reaches B at 12:50 p.m. with a velocity of 60 mi/h, what is the distance from A to B?

  6. At noon a car starts from rest at point A and proceeds with constant acceleration along a straight road toward point C, 35 miles away. If the constantly accelerated car arrives at C with a velocity of 60 mi/h, at what time does it arrive at C?

  7. River crossing If a=0.5 mi and v0=9 mi/h as in Example 4, what must the swimmer’s speed vS be in order that he drifts only 1 mile downstream as he crosses the river?

  8. River crossing Suppose that a=0.5 mi, v0=9 mi/h, and vS=3 mi/h as in Example 4, but that the velocity of the river is given by the fourth-degree function

    υR=υ0(1x4a4)

    rather than the quadratic function in Eq. (18). Now find how far downstream the swimmer drifts as he crosses the river.

  9. Interception of bomb A bomb is dropped from a helicopter hovering at an altitude of 800 feet above the ground. From the ground directly beneath the helicopter, a projectile is fired straight upward toward the bomb, exactly 2 seconds after the bomb is released. With what initial velocity should the projectile be fired in order to hit the bomb at an altitude of exactly 400 feet?

  10. Lunar lander A spacecraft is in free fall toward the surface of the moon at a speed of 1000 mph (mi/h). Its retrorockets, when fired, provide a constant deceleration of 20,000 mi/h2. At what height above the lunar surface should the astronauts fire the retrorockets to insure a soft touchdown? (As in Example 2, ignore the moon’s gravitational field.)

  11. Solar wind Arthur Clarke’s The Wind from the Sun (1963) describes Diana, a spacecraft propelled by the solar wind. Its aluminized sail provides it with a constant acceleration of 0.001g=0.0098 m/s2. Suppose this spacecraft starts from rest at time t=0 and simultaneously fires a projectile (straight ahead in the same direction) that travels at one-tenth of the speed c=3×108 m/s of light. How long will it take the spacecraft to catch up with the projectile, and how far will it have traveled by then?

  12. Length of skid A driver involved in an accident claims he was going only 25 mph. When police tested his car, they found that when its brakes were applied at 25 mph, the car skidded only 45 feet before coming to a stop. But the driver’s skid marks at the accident scene measured 210 feet. Assuming the same (constant) deceleration, determine the speed he was actually traveling just prior to the accident.

  13. Kinematic formula Use Eqs. (10) and (11) to show that v(t)2v02=2a[x(t)x0] for all t when the acceleration a=dv/dt is constant. Then use this “kinematic formula”—commonly presented in introductory physics courses—to confirm the result of Example 2.

1.3 Slope Fields and Solution Curves

Consider a differential equation of the form

(1)dydx=f(x, y)

where the right-hand function f(x,y) involves both the independent variable x and the dependent variable y. We might think of integrating both sides in (1) with respect to x, and hence write y(x)=f(x, y(x))dx+C. However, this approach does not lead to a solution of the differential equation, because the indicated integral involves the unknown function y(x) itself, and therefore cannot be evaluated explicitly. Actually, there exists no straightforward procedure by which a general differential equation can be solved explicitly. Indeed, the solutions of such a simple-looking differential equation as y=x2+y2 cannot be expressed in terms of the ordinary elementary functions studied in calculus textbooks. Nevertheless, the graphical and numerical methods of this and later sections can be used to construct approximate solutions of differential equations that suffice for many practical purposes.

Slope Fields and Graphical Solutions

There is a simple geometric way to think about solutions of a given differential equation y=f(x,y). At each point (x, y) of the xy-plane, the value of f(x,y) determines a slope m=f(x,y). A solution of the differential equation is simply a differentiable function whose graph y=y(x) has this “correct slope” at each point (x, y(x)) through which it passes—that is, y(x)=f(x,y(x)). Thus a solution curve of the differential equation y=f(x,y)—the graph of a solution of the equation—is simply a curve in the xy-plane whose tangent line at each point (x, y) has slope m=f(x,y). For instance, Fig. 1.3.1 shows a solution curve of the differential equation y=xy together with its tangent lines at three typical points.

FIGURE 1.3.1.

A solution curve for the differential equation y=xy together with tangent lines having

  • slope m1=x1y1 at the point (x1, y1);

  • slope m2=x2y2 at the point (x2, y2); and

  • slope m3=x3y3 at the point (x3, y3).

This geometric viewpoint suggests a graphical method for constructing approximate solutions of the differential equation y=f(x,y). Through each of a representative collection of points (x, y) in the plane we draw a short line segment having the proper slope m=f(x,y). All these line segments constitute a slope field (or a direction field) for the equation y=f(x,y).

Example 1

Figures 1.3.2 (a)(d) show slope fields and solution curves for the differential equation

(2)dydx=ky

with the values k=2, 0.5, 1, and 3 of the parameter k in Eq. (2). Note that each slope field yields important qualitative information about the set of all solutions of the differential equation. For instance, Figs. 1.3.2(a) and (b) suggest that each solution y(x) approaches ± as x+ if k>0, whereas Figs. 1.3.2(c) and (d) suggest that y(x)0 as x+ if k<0. Moreover, although the sign of k determines the direction of increase or decrease of y(x), its absolute value |k| appears to determine the rate of change of y(x). All this is apparent from slope fields like those in Fig. 1.3.2, even without knowing that the general solution of Eq. (2) is given explicitly by y(x)=Cekx.

A slope field suggests visually the general shapes of solution curves of the differential equation. Through each point a solution curve should proceed in such a direction that its tangent line is nearly parallel to the nearby line segments of the slope field. Starting at any initial point (a, b), we can attempt to sketch freehand an approximate solution curve that threads its way through the slope field, following the visible line segments as closely as possible.

FIGURE 1.3.2(a)

Slope field and solution curves for y=2y.

FIGURE 1.3.2(b)

Slope field and solution curves for y=(0.5)y.

FIGURE 1.3.2(c)

Slope field and solution curves for y=y.

FIGURE 1.3.2(d)

Slope field and solution curves for y=3y.

x/y 4 3 2 1 0 1 2 3 4
4 0 1 2 3 4 5 6 7 8
3 1 0 1 2 3 4 5 6 7
2 2 1 0 1 2 3 4 5 6
1 3 2 1 0 1 2 3 4 5
0 4 3 2 1 0 1 2 3 4
1 5 4 3 2 1 0 1 2 3
2 6 5 4 3 2 1 0 1 2
3 7 6 5 4 3 2 1 0 1
4 8 7 6 5 4 3 2 1 0

FIGURE 1.3.3.

Values of the slope y=xy for 4x, y4.

Example 2

Construct a slope field for the differential equation y=xy and use it to sketch an approximate solution curve that passes through the point (4,4).

Solution

Figure 1.3.3 shows a table of slopes for the given equation. The numerical slope m=xy appears at the intersection of the horizontal x-row and the vertical y-column of the table. If you inspect the pattern of upper-left to lower-right diagonals in this table, you can see that it was easily and quickly constructed. (Of course, a more complicated function f(x, y) on the right-hand side of the differential equation would necessitate more complicated calculations.) Figure 1.3.4 shows the corresponding slope field, and Fig. 1.3.5 shows an approximate solution curve sketched through the point (4,4) so as to follow this slope field as closely as possible. At each point it appears to proceed in the direction indicated by the nearby line segments of the slope field.

Although a spreadsheet program (for instance) readily constructs a table of slopes as in Fig. 1.3.3, it can be quite tedious to plot by hand a sufficient number of slope segments as in Fig. 1.3.4. However, most computer algebra systems include commands for quick and ready construction of slope fields with as many line segments as desired; such commands are illustrated in the application material for this section. The more line segments are constructed, the more accurately solution curves can be visualized and sketched. Figure 1.3.6 shows a “finer” slope field for the differential equation y=xy of Example 2, together with typical solution curves treading through this slope field.

FIGURE 1.3.4.

Slope field for y=xy corresponding to the table of slopes in Fig. 1.3.3.

FIGURE 1.3.5.

The solution curve through (4, 4).

If you look closely at Fig. 1.3.6, you may spot a solution curve that appears to be a straight line! Indeed, you can verify that the linear function y=x1 is a solution of the equation y=xy, and it appears likely that the other solution curves approach this straight line as an asymptote as x+. This inference illustrates the fact that a slope field can suggest tangible information about solutions that is not at all evident from the differential equation itself. Can you, by tracing the appropriate solution curve in this figure, infer that y(3)2 for the solution y(x) of the initial value problem y=xy, y(4)=4?

FIGURE 1.3.6.

Slope field and typical solution curves for y=xy.

Applications of Slope Fields

The next two examples illustrate the use of slope fields to glean useful information in physical situations that are modeled by differential equations. Example 3 is based on the fact that a baseball moving through the air at a moderate speed v (less than about 300 ft/s) encounters air resistance that is approximately proportional to v. If the baseball is thrown straight downward from the top of a tall building or from a hovering helicopter, then it experiences both the downward acceleration of gravity and an upward acceleration of air resistance. If the y-axis is directed downward, then the ball’s velocity v=dy/dt and its gravitational acceleration g=32 ft/s2 are both positive, while its acceleration due to air resistance is negative. Hence its total acceleration is of the form

(3)dvdt=gkv.

A typical value of the air resistance proportionality constant might be k=0.16.

Example 3

Falling baseball Suppose you throw a baseball straight downward from a helicopter hovering at an altitude of 3000 feet. You wonder whether someone standing on the ground below could conceivably catch it. In order to estimate the speed with which the ball will land, you can use your laptop’s computer algebra system to construct a slope field for the differential equation

FIGURE 1.3.7.

Slope field and typical solution curves for v=320.16v.

(4)dvdt=320.16v.

The result is shown in Fig. 1.3.7, together with a number of solution curves corresponding to different values of the initial velocity v(0) with which you might throw the baseball downward. Note that all these solution curves appear to approach the horizontal line v=200 as an asymptote. This implies that—however you throw it—the baseball should approach the limiting velocity v=200 ft/s instead of accelerating indefinitely (as it would in the absence of any air resistance). The handy fact that 60 mi/h=88 ft/s yields

v=200fts×60 mi/h88 ft/s136.36mih.

Perhaps a catcher accustomed to 100 mi/h fastballs would have some chance of fielding this speeding ball.

Comment

If the ball’s initial velocity is v(0)=200, then Eq. (4) gives v(0)=32(0.16)(200)=0, so the ball experiences no initial acceleration. Its velocity therefore remains unchanged, and hence v(t)200 is a constant “equilibrium solution” of the differential equation. If the initial velocity is greater than 200, then the initial acceleration given by Eq. (4) is negative, so the ball slows down as it falls. But if the initial velocity is less than 200, then the initial acceleration given by (4) is positive, so the ball speeds up as it falls. It therefore seems quite reasonable that, because of air resistance, the baseball will approach a limiting velocity of 200 ft/s—whatever initial velocity it starts with. You might like to verify that—in the absence of air resistance—this ball would hit the ground at over 300 mi/h.

In Section 2.1 we will discuss in detail the logistic differential equation

(5)dPdt=kP(MP)

that often is used to model a population P(t) that inhabits an environment with carrying capacity M. This means that M is the maximum population that this environment can sustain on a long-term basis (in terms of the maximum available food, for instance).

Example 4

Limiting population If we take k=0.0004 and M=150, then the logistic equation in (5) takes the form

(6)dPdt=0.0004P(150P)=0.06P0.0004P2.

FIGURE 1.3.8.

Slope field and typical solution curves for P=0.06P0.0004P2.

The positive term 0.06P on the right in (6) corresponds to natural growth at a 6%annual rate (with time t measured in years). The negative term 0.0004P2 represents the inhibition of growth due to limited resources in the environment.

Figure 1.3.8 shows a slope field for Eq. (6), together with a number of solution curves corresponding to possible different values of the initial population P(0). Note that all these solution curves appear to approach the horizontal line P=150 as an asymptote. This implies that—whatever the initial population—the population P(t) approaches the limiting population P=150 as t.

Comment

If the initial population is P(0)=150, then Eq. (6) gives

P(0)=0.0004(150)(150150)=0,

so the population experiences no initial (instantaneous) change. It therefore remains unchanged, and hence P(t)150 is a constant “equilibrium solution” of the differential equation. If the initial population is greater than 150, then the initial rate of change given by (6) is negative, so the population immediately begins to decrease. But if the initial population is less than 150, then the initial rate of change given by (6) is positive, so the population immediately begins to increase. It therefore seems quite reasonable to conclude that the population will approach a limiting value of 150—whatever the (positive) initial population.

Existence and Uniqueness of Solutions

Before one spends much time attempting to solve a given differential equation, it is wise to know that solutions actually exist. We may also want to know whether there is only one solution of the equation satisfying a given initial condition—that is, whether its solutions are unique.

Example 5

  1. Failure of existence The initial value problem

    (7)y=1x,y(0)=0

    has no solution, because no solution y(x)=(1/x)dx=ln |x|+C of the differential equation is defined at x=0. We see this graphically in Fig. 1.3.9, which shows a direction field and some typical solution curves for the equation y=1/x. It is apparent that the indicated direction field “forces” all solution curves near the y-axis to plunge downward so that none can pass through the point (0, 0).

    FIGURE 1.3.9.

    Direction field and typical solution curves for the equation y=1/x.

    FIGURE 1.3.10.

    Direction field and two different solution curves for the initial value problem y=2y, y(0)=0.

  2. Failure of uniqueness On the other hand, you can readily verify that the initial value problem

    (8)y=2y,y(0)=0

    has the two different solutions y1(x)=x2 and y2(x)0 (see Problem 27). Figure 1.3.10 shows a direction field and these two different solution curves for the initial value problem in (8). We see that the curve y1(x)=x2 threads its way through the indicated direction field, whereas the differential equation y=2y specifies slope y=0 along the x-axis y2(x)=0.

Example 5 illustrates the fact that, before we can speak of “the” solution of an initial value problem, we need to know that it has one and only one solution. Questions of existence and uniqueness of solutions also bear on the process of mathematical modeling. Suppose that we are studying a physical system whose behavior is completely determined by certain initial conditions, but that our proposed mathematical model involves a differential equation not having a unique solution satisfying those conditions. This raises an immediate question as to whether the mathematical model adequately represents the physical system.

The theorem stated below implies that the initial value problem y=f(x,y), y(a)=b has one and only one solution defined near the point x=a on the x-axis, provided that both the function f and its partial derivative f/y are continuous near the point (a, b) in the xy-plane. Methods of proving existence and uniqueness theorems are discussed in the Appendix.

FIGURE 1.3.11.

The rectangle R and x-interval I of Theorem 1, and the solution curve y=y(x) through the point (a, b).

Remark 1

In the case of the differential equation dy/dx=y of Example 1 and Fig. 1.3.2(c), both the function f(x,y)=y and the partial derivative f/y=1 are continuous everywhere, so Theorem 1 implies the existence of a unique solution for any initial data (a, b). Although the theorem ensures existence only on some open interval containing x=a, each solution y(x)=Cex actually is defined for all x.

Remark 2

In the case of the differential equation dy/dx=2y of Example 5(b) and Eq. (8), the function f(x,y)=2y is continuous wherever y>0, but the partial derivative f/y=1/y is discontinuous when y=0, and hence at the point (0, 0). This is why it is possible for there to exist two different solutions y1(x)=x2 and y2(x)0, each of which satisfies the initial condition y(0)=0.

Remark 3

In Example 7 of Section 1.1 we examined the especially simple differential equation dy/dx=y2. Here we have f(x,y)=y2 and f/y=2y. Both of these functions are continuous everywhere in the xy-plane, and in particular on the rectangle 2<x<2, 0<y<2. Because the point (0, 1) lies in the interior of this rectangle, Theorem 1 guarantees a unique solution—necessarily a continuous function—of the initial value problem

(10)dydx=y2,y(0)=1

on some open x-interval containing a=0. Indeed this is the solution

y(x)=11x

that we discussed in Example 7. But y(x)=1/(1x) is discontinuous at x=1, so our unique continuous solution does not exist on the entire interval 2<x<2. Thus the solution interval I of Theorem 1 may not be as wide as the rectangle R where f and f/y are continuous. Geometrically, the reason is that the solution curve provided by the theorem may leave the rectangle—wherein solutions of the differential equation are guaranteed to exist—before it reaches the one or both ends of the interval (see Fig. 1.3.12).

FIGURE 1.3.12.

The solution curve through the initial point (0, 1) leaves the rectangle R before it reaches the right side of R.

The following example shows that, if the function f(x,y) and/or its partial derivative f/y fail to satisfy the continuity hypothesis of Theorem 1, then the initial value problem in (9) may have either no solution or many—even infinitely many—solutions.

Example 6

Consider the first-order differential equation

(11)xdydx=2y.

Applying Theorem 1 with f(x,y)=2y/x and f/y=2/x, we conclude that Eq. (11) must have a unique solution near any point in the xy-plane where x0. Indeed, we see immediately by substitution in (11) that

(12)y(x)=Cx2

satisfies Eq. (11) for any value of the constant C and for all values of the variable x. In particular, the initial value problem

FIGURE 1.3.13.

There are infinitely many solution curves through the point (0, 0), but no solution curves through the point (0, b) if b0.

(13)xdydx=2y,y(0)=0

has infinitely many different solutions, whose solution curves are the parabolas y=Cx2 illustrated in Fig. 1.3.13. (In case C=0 the “parabola” is actually the x-axis y=0.)

Observe that all these parabolas pass through the origin (0, 0), but none of them passes through any other point on the y-axis. It follows that the initial value problem in (13) has infinitely many solutions, but the initial value problem

(14)xdydx=2y,y(0)=b

has no solution if b0.

Finally, note that through any point off the y-axis there passes only one of the parabolas y=Cx2. Hence, if a0, then the initial value problem

(15)xdydx=2y,y(a)=b

has a unique solution on any interval that contains the point x=a but not the origin x=0. In summary, the initial value problem in (15) has

  • a unique solution near (a, b) if a0;

  • no solution if a=0 but b0;

  • infinitely many solutions if a=b=0.

Still more can be said about the initial value problem in (15). Consider a typical initial point off the y-axis—for instance the point (1,1) indicated in Fig. 1.3.14. Then for any value of the constant C the function defined by

(16)y(x)={x2if x0,Cx2if x>0

is continuous and satisfies the initial value problem

(17)xdydx=2y,y(1)=1.

FIGURE 1.3.14.

There are infinitely many solution curves through the point (1, 1).

For a particular value of C, the solution curve defined by (16) consists of the left half of the parabola y=x2 and the right half of the parabola y=Cx2. Thus the unique solution curve near (1,1) branches at the origin into the infinitely many solution curves illustrated in Fig. 1.3.14.

We therefore see that Theorem 1 (if its hypotheses are satisfied) guarantees uniqueness of the solution near the initial point (a, b), but a solution curve through (a, b) may eventually branch elsewhere so that uniqueness is lost. Thus a solution may exist on a larger interval than one on which the solution is unique. For instance, the solution y(x)=x2 of the initial value problem in (17) exists on the whole x-axis, but this solution is unique only on the negative x-axis <x<0.

1.3 Problems

In Problems 1 through 10, we have provided the slope field of the indicated differential equation, together with one or more solution curves. Sketch likely solution curves through the additional points marked in each slope field.

  1. dydx=ysin x

    FIGURE 1.3.15.

  2. dydx=x+y

    FIGURE 1.3.16.

  3. dydx=ysin x

    FIGURE 1.3.17.

  4. dydx=xy

    FIGURE 1.3.18.

  5. dydx=yx+1

    FIGURE 1.3.19.

  6. dydx=xy+1

    FIGURE 1.3.20.

  7. dydx=sin x+sin y

    FIGURE 1.3.21.

  8. dydx=x2y

    FIGURE 1.3.22.

  9. dydx=x2y2

    FIGURE 1.3.23.

  10. dydx=x2+sin y

    FIGURE 1.3.24.

A more detailed version of Theorem 1 says that, if the function f(x,y) is continuous near the point (a, b), then at least one solution of the differential equation y=f(x,y) exists on some open interval I containing the point x=a and, moreover, that if in addition the partial derivative f/y is continuous near (a, b), then this solution is unique on some (perhaps smaller) interval J. In Problems 11 through 20, determine whether existence of at least one solution of the given initial value problem is thereby guaranteed and, if so, whether uniqueness of that solution is guaranteed.

  1. dydx=2x2y2;y(1)=1

  2. dydx=x ln  y;y(1)=1

  3. dydx=y3;y(0)=1

  4. dydx=y3;y(0)=0

  5. dydx=xy;y(2)=2

  6. dydx=xy;y(2)=1

  7. ydydx=x1;y(0)=1

  8. ydydx=x1;y(1)=0

  9. dydx=ln (1+y2);y(0)=0

  10. dydx=x2y2;y(0)=1

In Problems 21 and 22, first use the method of Example 2 to construct a slope field for the given differential equation. Then sketch the solution curve corresponding to the given initial condition. Finally, use this solution curve to estimate the desired value of the solution y(x).

  1. y=x+y,   y(0)=0;   y(4)=?

  2. y=yx,   y(4)=0;   y(4)=?

Problems 23 and 24 are like Problems 21 and 22, but now use a computer algebra system to plot and print out a slope field for the given differential equation. If you wish (and know how), you can check your manually sketched solution curve by plotting it with the computer.

  1. y=x2+y21,   y(0)=0;   y(2)=?

  2. y=x+12y2,   y(2)=0;   y(2)=?

  3. Falling Parachutist You bail out of the helicopter of Example 3 and pull the ripcord of your parachute. Now k=1.6 in Eq. (3), so your downward velocity satisfies the initial value problem

    dvdt=321.6v,v(0)=0.

    In order to investigate your chances of survival, construct a slope field for this differential equation and sketch the appropriate solution curve. What will your limiting velocity be? Will a strategically located haystack do any good? How long will it take you to reach 95% of your limiting velocity?

  4. Deer Population Suppose the deer population P(t) in a small forest satisfies the logistic equation

    dPdt=0.0225P0.0003P2.

    Construct a slope field and appropriate solution curve to answer the following questions: If there are 25 deer at time t=0 and t is measured in months, how long will it take the number of deer to double? What will be the limiting deer population?

The next seven problems illustrate the fact that, if the hypotheses of Theorem 1 are not satisfied, then the initial value problem y=f(x,y), y(a)=b may have either no solutions, finitely many solutions, or infinitely many solutions.

    1. Verify that if c is a constant, then the function defined piecewise by

      y(x)={0for xc,(xc)2for x>c

      satisfies the differential equation y=2y for all x (including the point x=c). Construct a figure illustrating the fact that the initial value problem y=2y, y(0)=0 has infinitely many different solutions.

    2. For what values of b does the initial value problem y=2y, y(0)=b have (i) no solution, (ii) a unique solution that is defined for all x?

  1. Verify that if k is a constant, then the function y(x)kx satisfies the differential equation xy=y for all x. Construct a slope field and several of these straight line solution curves. Then determine (in terms of a and b) how many different solutions the initial value problem xy=y, y(a)=b has—one, none, or infinitely many.

  2. Verify that if c is a constant, then the function defined piecewise by

    y(x)={0for xc,(xc)3for x>c

    satisfies the differential equation y=3y2/3 for all x. Can you also use the “left half” of the cubic y=(xc)3 in piecing together a solution curve of the differential equation? (See Fig. 1.3.25.) Sketch a variety of such solution curves. Is there a point (a, b) of the xy-plane such that the initial value problem y=3y2/3, y(a)=b has either no solution or a unique solution that is defined for all x? Reconcile your answer with Theorem 1.

    FIGURE 1.3.25.

    A suggestion for Problem 29.

  3. Verify that if c is a constant, then the function defined piecewise by

    y(x)={+1if xc,cos(xc)if c<x<c+π,1if xc+π

    satisfies the differential equation y=1y2 for all x. (Perhaps a preliminary sketch with c=0 will be helpful.) Sketch a variety of such solution curves. Then determine (in terms of a and b) how many different solutions the initial value problem y=1y2, y(a)=b has.

  4. Carry out an investigation similar to that in Problem 30, except with the differential equation y=+1y2. Does it suffice simply to replace cos(xc) with sin(xc) in piecing together a solution that is defined for all x?

  5. Verify that if c>0, then the function defined piecewise by

    y(x)={0if x2c,(x2c)2if x2>c

    satisfies the differential equation y=4xy for all x. Sketch a variety of such solution curves for different values of c. Then determine (in terms of a and b) how many different solutions the initial value problem y=4xy, y(a)=b has.

  6. If c0, verify that the function defined by y(x)=x/(cx1) (with the graph illustrated in Fig. 1.3.26) satisfies the differential equation x2y+y2=0 if x1/c. Sketch a variety of such solution curves for different values of c. Also, note the constant-valued function y(x)0 that does not result from any choice of the constant c. Finally, determine (in terms of a and b) how many different solutions the initial value problem x2y+y2=0, y(a)=b has.

    FIGURE 1.3.26.

    Slope field for x2y+y2=0 and graph of a solution y(x)=x/(cx1).

    1. Use the direction field of Problem 5 to estimate the values at x=1 of the two solutions of the differential equation y=yx+1 with initial values y(1)=1.2 and y(1)=0.8.

    2. Use a computer algebra system to estimate the values at x=3 of the two solutions of this differential equation with initial values y(3)=3.01 and y(3)=2.99.

    The lesson of this problem is that small changes in initial conditions can make big differences in results.

    1. Use the direction field of Problem 6 to estimate the values at x=2 of the two solutions of the differential equation y=xy+1 with initial values y(3)=0.2 and y(3)=+0.2.

    2. Use a computer algebra system to estimate the values at x=2 of the two solutions of this differential equation with initial values y(3)=0.5 and y(3)=+0.5.

    The lesson of this problem is that big changes in initial conditions may make only small differences in results.

1.3 Application Computer-Generated Slope Fields and Solution Curves

Widely available computer algebra systems and technical computing environments include facilities to automate the construction of slope fields and solution curves, as do some graphing calculators (see Figs. 1.3.2729).

The Expanded Applications site at the URL indicated in the margin includes a discussion of Maple™, Mathematica™, and Matlab™ resources for the investigation of differential equations. For instance, the Maple command

with(DEtools):
DEplot(diff(y(x),x)=sin(x-y(x)), y(x), x=-5..5, y=-5..5);

and the Mathematica command

VectorPlot[{1, Sin[x-y]},{x, -5, 5}, {y, -5, 5}]

produce slope fields similar to the one shown in Fig. 1.3.29. Figure 1.3.29 itself was generated with the Matlab program dfield [John Polking and David Arnold, Ordinary Differential Equations Using Matlab, 3rd edition, Hoboken, NJ: Pearson, 2003] that is freely available for educational use (math.rice.edu/~dfield). This web site also provides a stand-alone Java version of that can be used in a web browser. When a differential equation is entered in the dfield setup menu (Fig. 1.3.30), you can (with mouse button clicks) plot both a slope field and the solution curve (or curves) through any desired point (or points). Another freely available and user-friendly Matlab-based ODE package with impressive graphical capabilities is Iode (www.math.uiuc.edu/iode).

FIGURE 1.3.27.

TI-84 Plus CE™ graphing calculator and TI-Nspire™ CX CAS handheld. Screenshot from Texas Instruments Incorporated. Courtesy of Texas Instruments Incorporated.

Modern technology platforms offer even further interactivity by allowing the user to vary initial conditions and other parameters “in real time.” Mathematica’s Manipulate command was used to generate Fig. 1.3.31, which shows three particular solutions of the differential equation dy/dx=sin(xy). The solid curve corresponds to the initial condition y(1)=0. As the “locator point” initially at (1, 0) is dragged—by mouse or touchpad—to the point (0, 3) or (2,2), the solution curve immediately follows, resulting in the dashed curves shown. The TI-Nspire CX CAS has similar capability; indeed, as Fig. 1.3.28 appears on the Nspire display, each of the initial points (0,b) can be dragged to different locations using the Nspire’s touchpad, with the corresponding solution curves being instantly redrawn.

FIGURE 1.3.28.

Slope field and solution curves for the differential equation

dydx=sin(xy)

with initial points (0, b), b=3, 1, 2, 0, 2, 4 and window 5x, y5 on a TI-89 graphing calculator.

FIGURE 1.3.29.

Computer-generated slope field and solution curves for the differential equation y=sin(xy).

FIGURE 1.3.30.

Matlab dfield setup to construct slope field and solution curves for y=sin(xy).

FIGURE 1.3.31.

Interactive Mathematica solution of the differential equation y=sin(xy). The “locator point” corresponding to the initial condition y(1)=0 can be dragged to any other point in the display, causing the solution curve to be automatically redrawn.

Use a graphing calculator or computer system in the following investigations. You might warm up by generating the slope fields and some solution curves for Problems 1 through 10 in this section.

Investigation A: Plot a slope field and typical solution curves for the differential equation dy/dx=sin(xy), but with a larger window than that of Fig. 1.3.29. With 10x10, 10y10, for instance, a number of apparent straight line solution curves should be visible, especially if your display allows you to drag the initial point interactively from upper left to lower right.

  1. Substitute y=ax+b in the differential equation to determine what the coefficients a and b must be in order to get a solution. Are the results consistent with what you see on the display?

  2. A computer algebra system gives the general solution

    y(x)=x2 tan1(x2CxC).

    Plot this solution with selected values of the constant C to compare the resulting solution curves with those indicated in Fig. 1.3.28. Can you see that no value of C yields the linear solution y=xπ/2 corresponding to the initial condition y(π/2)=0? Are there any values of C for which the corresponding solution curves lie close to this straight line solution curve?

Investigation B: For your own personal investigation, let n be the smallest digit in your student ID number that is greater than 1, and consider the differential equation

dydx=1ncos(xny).
  1. First investigate (as in part (a) of Investigation A) the possibility of straight line solutions.

  2. Then generate a slope field for this differential equation, with the viewing window chosen so that you can picture some of these straight lines, plus a sufficient number of nonlinear solution curves that you can formulate a conjecture about what happens to y(x) as x+. State your inference as plainly as you can. Given the initial value y(0)=y0, try to predict (perhaps in terms of y0) how y(x) behaves as x+.

  3. A computer algebra system gives the general solution

    y(x)=1n[x+2 tan1(1xC)].

    Can you make a connection between this symbolic solution and your graphically generated solution curves (straight lines or otherwise)?

1.4 Separable Equations and Applications

In the preceding sections we saw that if the function f(x, y) does not involve the variable y, then solving the first-order differential equation

(1)dydx=f(x,y)

is a matter of simply finding an antiderivative. For example, the general solution of

(2)dydx=6x

is given by

y(x)=6xdx=3x2+C.

If instead f(x, y) does involve the dependent variable y, then we can no longer solve the equation merely by integrating both sides: The differential equation

(3a)dydx=6xy

differs from Eq. (2) only in the factor y appearing on the right-hand side, but this is enough to prevent us from using the same approach to solve Eq. (3a) that was successful with Eq. (2).

And yet, as we will see throughout the remainder of this chapter, differential equations like (3a) often can, in fact, be solved by methods which are based on the idea of “integrating both sides.” The idea behind these techniques is to rewrite the given equation in a form that, while equivalent to the given equation, allows both sides to be integrated directly, thus leading to the solution of the original differential equation.

The most basic of these methods, separation of variables, can be applied to Eq. (3a). First, we note that the right-hand function f(x,y)=6xy can be viewed as the product of two expressions, one involving only the independent variable x, and the other involving only the dependent variable y:

(3b)dydx=(6x)y.

Next, we informally break up the derivative dy/dx into the “free-floating” differentials dx and dy—a notational convenience that leads to correct results, as we will see below—and then multiply by dx and divide by y in Eq. (3b), leading to

(3c)dyy=6x dx.

Equation (3c) is an equivalent version of the original differential equation in (3a), but with the variables x and y separated (that is, by the equal sign), and this is what allows us to integrate both sides. The left-hand side is integrated with respect to y (with no “interference” from the variable x), and vice versa for the right-hand side. This leads to

dyy=6x dx,

or

(4)ln |y|=3x2+C.

This gives the general solution of Eq. (3a) implicitly, and a family of solution curves is shown in Fig. 1.4.1.

In this particular case we can go on to solve for y to give the explicit general solution

(5)y(x)=±e3x2+C=±e3x2eC=Ae3x2,

FIGURE 1.4.1.

Slope field and solution curves for y=6xy.

where A represents the constant ±eC, which can take on any nonzero value. If we impose an initial condition on Eq. (3a), say y(0)=7, then in Eq. (5) we find that A=7, yielding the particular solution

y(x)=7e3x2,

which is the upper emphasized solution curve shown in Fig. 1.4.1. In the same way, the initial condition y(0)=4 leads to the particular solution

y(x)=4e3x2,

which is the lower emphasized solution curve shown in Fig. 1.4.1.

To complete this example, we note that whereas the constant A in Eq. (5) is nonzero, taking A=0 in (5) leads to y(x)0, and this is, in fact, a solution of the given differential equation (3a). Thus Eq. (5) actually provides a solution of (3a) for all values of the constant A, including A=0. Why did the method of separation of variables fail to capture all solutions of Eq. (3a)? The reason is that in the step in which we actually separated the variables, that is, in passing from Eq. (3b) to (3c), we divided by y, thus (tacitly) assuming that y0. As a result, our general solution (5), with its restriction that A0, “missed” the particular solution y(x)0 corresponding to A=0. Such solutions are known as singular solutions, and we say more about them—together with implicit and general solutions—below.

In general, the first-order differential equation (1) is called separable provided that f(x,y) can be written as the product of a function of x and a function of y:

dydx=f(x, y)=g(x)k(y)=g(x)h(y),

where h(y)=1/k(y). In this case the variables x and y can be separated—isolated on opposite sides of an equation—by writing informally the equation

h(y)dy=g(x)dx,

which we understand to be concise notation for the differential equation

(6)h(y)dydx=g(x).

(In the preceding example, h(y)=1y and g(x)=6x.) As illustrated above, we can solve this type of differential equation simply by integrating both sides with respect to x:

h(y(x))dydxdx=g(x)dx+C;

equivalently,

(7)h(y)dy=g(x)dx+C.

All that is required is that the antiderivatives

H(y)=h(y)dy  and  G(x)=g(x)dx

can be found. To see that Eqs. (6) and (7) are equivalent, note the following consequence of the chain rule:

Dx[H(y(x))]=H(y(x))y(x)=h(y)dydx=g(x)=Dx[G(x)],

which in turn is equivalent to

(8)H(y(x))=G(x)+C,

because two functions have the same derivative on an interval if and only if they differ by a constant on that interval.

Example 1

Solve the differential equation

(9)dydx=42x3y25.

Solution

Because

42x3y25=(42x)·13y25=g(x)k(y)

is the product of a function that depends only on x, and one that depends only on y, Eq. (9) is separable, and thus we can proceed in much the same way as in Eq. (3a). Before doing so, however, we note an important feature of Eq. (9) not shared by Eq. (3a): The function k(y)=13y25 is not defined for all values of y. Indeed, setting 3y25 equal to zero shows that k(y), and thus dydx itself, becomes infinite as y approaches either of ±53. Because an infinite slope corresponds to a vertical line segment, we would therefore expect the line segments in the slope field for this differential equation to be “standing straight up” along the two horizontal lines y=±53±1.29; as Fig. 1.4.2 shows (where these two lines are dashed), this is indeed what we find.

FIGURE 1.4.2.

Slope field and solution curves for y=(42x)/(3y25) in Example 1.

What this means for the differential equation (9) is that no solution curve of this equation can cross either of the horizontal lines y=±53, simply because along these lines dydx is undefined. Effectively, then, these lines divide the plane into three regions—defined by the conditions y>53, 53<y<53, and y<53—with all solution curves of Eq. (9) remaining confined to one of these regions.

With this in mind, the general solution of the differential equation in Eq. (9) is easy to find, at least in implicit form. Separating variables and integrating both sides leads to

3y25 dy=42x dx,

and thus

(10)y35y=4xx2+C.

Note that unlike Eq. (4), the general solution in Eq. (10) cannot readily be solved for y; thus we cannot directly plot the solution curves of Eq. (9) in the form y=y(x), as we would like. However, what we can do is rearrange Eq. (10) so that the constant C is isolated on the right-hand side:

(11)y35y(4xx2)=C.

This shows that the solution curves of the differential equation in Eq. (9) are contained in the level curves (also known as contours) of the function

(12)F(x, y)=y35y(4xx2).

Because no particular solution curve of Eq. (9) can cross either of the lines y=±53—despite the fact that the level curves of F(x,y) freely do so—the particular solution curves of Eq. (9) are those portions of the level curves of F(x,y) which avoid the lines y=±53.

For example, suppose that we wish to solve the initial value problem

(13)dydx=42x3y25,y(1)=3.

Substituting x=1 and y=3 into our general solution (10) gives C=9. Therefore our desired solution curve lies on the level curve

(14)y35y(4xx2)=9

of F(x, y); Fig. 1.4.2 shows this and other level curves of F(x, y). However, because the solution curve of the initial value problem (13) must pass through the point (1, 3), which lies above the line y=53 in the xy-plane, our desired solution curve is restricted to that portion of the level curve (14) which satisfies y>53. (In Fig. 1.4.2 the solution curve of the initial value problem (13) is drawn more heavily than the remainder of the level curve (14).) In the same way, Figure 1.4.2 also shows the particular solutions of Eq. (9) subject to the initial conditions y(1)=0 and y(1)=2. In each of these cases, the curve corresponding to the desired particular solution is only a piece of a larger level curve of the function F(x, y). (Note that in fact, some of the level curves of F themselves consist of two pieces.)

FIGURE 1.4.3.

Graph of f(y)=y35y9.

Finally, despite the difficulty of solving Eq. (14) for y by algebraic means, we can nonetheless “solve” for y in the sense that, when a specific value of x is substituted in (14), we can attempt to solve numerically for y. For instance, taking x=4 yields the equation

f(y)=y35y9=0;

Fig. 1.4.3 shows the graph of f. Using technology we can solve for the single real root y2.8552, thus yielding the value y(4)2.8552 for the solution of the initial value problem (13). By repeating this process for other values of x, we can create a table (like the one shown below) of corresponding x- and y-values for the solution of (13); such a table serves effectively as a “numerical solution” of this initial value problem.

x 1 0 1 2 3 4 5 6
y 2.5616 2.8552 3 3.0446 3 2.8552 2.5616 1.8342

Implicit, General, and Singular Solutions

The equation K(x,y)=0 is commonly called an implicit solution of a differential equation if it is satisfied (on some interval) by some solution y=y(x) of the differential equation. But note that a particular solution y=y(x) of K(x,y)=0 may or may not satisfy a given initial condition. For example, differentiation of x2+y2=4 yields

x+ydydx=0,

so x2+y2=4 is an implicit solution of the differential equation x+yy=0. But only the first of the two explicit solutions

y(x)=+4x2andy(x)=4x2

satisfies the initial condition y(0)=2 (Fig. 1.4.4).

Remark 1

You should not assume that every possible algebraic solution y=y(x) of an implicit solution satisfies the same differential equation. For instance, if we multiply the implicit solution x2+y24=0 by the factor (y2x), then we get the new implicit solution

(y2x)(x2+y24)=0

that yields (or “contains”) not only the previously noted explicit solutions y=+4x2 and y=4x2 of the differential equation x+yy=0, but also the additional function y=2x that does not satisfy this differential equation.

FIGURE 1.4.4.

Slope field and solution curves for y=x/y.

Remark 2

Similarly, solutions of a given differential equation can be either gained or lost when it is multiplied or divided by an algebraic factor. For instance, consider the differential equation

(15)(y2x)ydydx=x(y2x)

having the obvious solution y=2x. But if we divide both sides by the common factor (y2x), then we get the previously discussed differential equation

(16)ydydx=x,orx+ydydx=0,

of which y=2x is not a solution. Thus we “lose” the solution y=2x of Eq. (15) upon its division by the factor (y2x); alternatively, we “gain” this new solution when we multiply Eq. (16) by (y2x). Such elementary algebraic operations to simplify a given differential equation before attempting to solve it are common in practice, but the possibility of loss or gain of such “extraneous solutions” should be kept in mind.

A solution of a differential equation that contains an “arbitrary constant” (like the constant C appearing in Eqs. (4) and (10)) is commonly called a general solution of the differential equation; any particular choice of a specific value for C yields a single particular solution of the equation.

The argument preceding Example 1 actually suffices to show that every particular solution of the differential equation h(y)y=g(x) in (6) satisfies the equation H(y(x))=G(x)+C in (8). Consequently, it is appropriate to call (8) not merely a general solution of (6), but the general solution of (6).

FIGURE 1.4.5.

The general solution curves y=(xC)2 and the singular solution curve y=0 of the differential equation (y)2=4y.

In Section 1.5 we shall see that every particular solution of a linear first-order differential equation is contained in its general solution. By contrast, it is common for a nonlinear first-order differential equation to have both a general solution involving an arbitrary constant C and one or several particular solutions that cannot be obtained by selecting a value for C. These exceptional solutions are frequently called singular solutions. In Problem 30 we ask you to show that the general solution of the differential equation (y)2=4y yields the family of parabolas y=(xC)2 illustrated in Fig. 1.4.5, and to observe that the constant-valued function y(x)0 is a singular solution that cannot be obtained from the general solution by any choice of the arbitrary constant C.

Example 2

Find all solutions of the differential equation

dydx=6x(y1)2/3.

Solution

Separation of variables gives

13(y1)2/3dy=2x dx;(y1)1/3=x2+C;y(x)=1+(x2+c)3.

Positive values of the arbitrary constant C give the solution curves in Fig. 1.4.6 that lie above the line y=1, whereas negative values yield those that dip below it. The value C=0 gives the solution y(x)=1+x6, but no value of C gives the singular solution y(x)1 that was lost when the variables were separated. Note that the two different solutions y(x)1 and y(x)=1+(x21)3 both satisfy the initial condition y(1)=1. Indeed, the whole singular solution curve y=1 consists of points where the solution is not unique and where the function f(x,y)=6x(y1)2/3 is not differentiable.

FIGURE 1.4.6.

General and singular solution curves for y=6x(y1)2/3.

Natural Growth and Decay

The differential equation

(17)dxdt=kx(k  a constant)

serves as a mathematical model for a remarkably wide range of natural phenomena—any involving a quantity whose time rate of change is proportional to its current size. Here are some examples.

Population Growth: Suppose that P(t) is the number of individuals in a population (of humans, or insects, or bacteria) having constant birth and death rates β and δ (in births or deaths per individual per unit of time). Then, during a short time interval Δt, approximately βP(t)Δt births and δP(t)Δt deaths occur, so the change in P(t) is given approximately by

ΔP(βδ)P(t)Δt,

and therefore

(18)dPdt=limΔt0ΔPΔt=kP,

where k=βδ.

Compound Interest: Let A(t) be the number of dollars in a savings account at time t (in years), and suppose that the interest is compounded continuously at an annual interest rate r. (Note that 10% annual interest means that r=0.10.) Continuous compounding means that during a short time interval Δt, the amount of interest added to the account is approximately ΔA=rA(t)Δt, so that

(19)dAdt=limΔt0ΔAΔt=rA,

Radioactive Decay: Consider a sample of material that contains N(t) atoms of a certain radioactive isotope at time t. It has been observed that a constant fraction of those radioactive atoms will spontaneously decay (into atoms of another element or into another isotope of the same element) during each unit of time. Consequently, the sample behaves exactly like a population with a constant death rate and no births. To write a model for N(t), we use Eq. (18) with N in place of P, with k>0 in place of δ, and with β=0. We thus get the differential equation

(20)dNdt=kN.

The value of k depends on the particular radioactive isotope.

The key to the method of radiocarbon dating is that a constant proportion of the carbon atoms in any living creature is made up of the radioactive isotope C14 of carbon. This proportion remains constant because the fraction of C14 in the atmosphere remains almost constant, and living matter is continuously taking up carbon from the air or is consuming other living matter containing the same constant ratio of C14 atoms to ordinary C12 atoms. This same ratio permeates all life, because organic processes seem to make no distinction between the two isotopes.

The ratio of C14 to normal carbon remains constant in the atmosphere because, although C14 is radioactive and slowly decays, the amount is continuously replenished through the conversion of N14 (ordinary nitrogen) to C14 by cosmic rays bombarding the upper atmosphere. Over the long history of the planet, this decay and replenishment process has come into nearly steady state.

Of course, when a living organism dies, it ceases its metabolism of carbon and the process of radioactive decay begins to deplete its C14 content. There is no replenishment of this C14, and consequently the ratio of C14 to normal carbon begins to drop. By measuring this ratio, the amount of time elapsed since the death of the organism can be estimated. For such purposes it is necessary to measure the decay constant k. For C14, it is known that k0.0001216 if t is measured in years.

(Matters are not as simple as we have made them appear. In applying the technique of radiocarbon dating, extreme care must be taken to avoid contaminating the sample with organic matter or even with ordinary fresh air. In addition, the cosmic ray levels apparently have not been constant, so the ratio of C14 in the atmosphere has varied over the past centuries. By using independent methods of dating samples, researchers in this area have compiled tables of correction factors to enhance the accuracy of this process.)

Drug Elimination: In many cases the amount A(t) of a certain drug in the bloodstream, measured by the excess over the natural level of the drug, will decline at a rate proportional to the current excess amount. That is,

(21)dAdt=λA,

where λ>0. The parameter λ is called the elimination constant of the drug.

The Natural Growth Equation

The prototype differential equation dx/dt=kx with x(t)>0 and k a constant (either negative or positive) is readily solved by separating the variables and integrating:

1xdx=kdt;ln x=kt+C.

FIGURE 1.4.7.

Natural growth.

Then we solve for x:

eln x=ekt+C;x=x(t)=eCekt=Aekt.

Because C is a constant, so is A=eC. It is also clear that A=x(0)=x0, so the particular solution of Eq. (17) with the initial condition x(0)=x0 is simply

(22)x(t)=x0ekt.

Because of the presence of the natural exponential function in its solution, the differential equation

(23)dxdt=kx

is often called the exponential or natural growth equation. Figure 1.4.7 shows a typical graph of x(t) in the case k>0; the case k<0 is illustrated in Fig. 1.4.8.

FIGURE 1.4.8.

Natural decay.

Example 3

World population According to data listed at www.census.gov, the world’s total population reached 6 billion persons in mid-1999, and was then increasing at the rate of about 212 thousand persons each day. Assuming that natural population growth at this rate continues, we want to answer these questions:

  1. What is the annual growth rate k?

  2. What will be the world population at the middle of the 21st century?

  3. How long will it take the world population to increase tenfold—thereby reaching the 60 billion that some demographers believe to be the maximum for which the planet can provide adequate food supplies?

Solution

  1. We measure the world population P(t) in billions and measure time in years. We take t=0 to correspond to (mid) 1999, so P0=6. The fact that P is increasing by 212,000, or 0.000212 billion, persons per day at time t=0 means that

    P(0)=(0.000212)(365.25)0.07743

    billion per year. From the natural growth equation P=kP with t=0 we now obtain

    k=P(0)P(0)0.0774360.0129.

    Thus the world population was growing at the rate of about 1.29% annually in 1999. This value of k gives the world population function

    P(t)=6e0.0129t.
  2. With t=51 we obtain the prediction

    P(51)=6e(0.0129)(51)11.58 (billion)

    for the world population in mid-2050 (so the population will almost have doubled in the just over a half-century since 1999).

  3. The world population should reach 60 billion when

    60=6e0.0129t;that is, when t = ln 100.0129 178;

    and thus in the year 2177.

Note

Actually, the rate of growth of the world population is expected to slow somewhat during the next half-century, and the best current prediction for the 2050 population is “only” 9.1 billion. A simple mathematical model cannot be expected to mirror precisely the complexity of the real world.

The decay constant of a radioactive isotope is often specified in terms of another empirical constant, the half-life of the isotope, because this parameter is more convenient. The half-life τ of a radioactive isotope is the time required for half of it to decay. To find the relationship between k and τ, we set t=τ and N=12N0 in the equation N(t)=N0ekt, so that 12N0=N0ekτ. When we solve for τ, we find that

(24)τ=ln 2k.

For example, the half-life of C14 is τ(ln 2)/(0.0001216), approximately 5700 years.

Example 4

Radiometric dating A specimen of charcoal found at Stonehenge turns out to contain 63% as much C14 as a sample of present-day charcoal of equal mass. What is the age of the sample?

Solution

We take t=0 as the time of the death of the tree from which the Stonehenge charcoal was made and N0 as the number of C14 atoms that the Stonehenge sample contained then. We are given that N=(0.63)N0 now, so we solve the equation (0.63)N0=N0ekt with the value k=0.0001216. Thus we find that

t=ln(0.63)0.00012163800 (years).

Thus the sample is about 3800 years old. If it has any connection with the builders of Stonehenge, our computations suggest that this observatory, monument, or temple—whichever it may be—dates from 1800 b.c. or earlier.

Cooling and Heating

According to Newton’s law of cooling (Eq. (3) of Section 1.1), the time rate of change of the temperature T(t) of a body immersed in a medium of constant temperature A is proportional to the difference AT. That is,

(25)dTdt=k(AT),

where k is a positive constant. This is an instance of the linear first-order differential equation with constant coefficients:

(26)dxdt=ax+b.

It includes the exponential equation as a special case (b=0) and is also easy to solve by separation of variables.

Example 5

Cooling A 4-lb roast, initially at 50°F, is placed in a 375°F oven at 5:00 p.m. After 75 minutes it is found that the temperature T(t) of the roast is 125°F. When will the roast be 150°F (medium rare)?

Solution

We take time t in minutes, with t=0 corresponding to 5:00 p.m. We also assume (somewhat unrealistically) that at any instant the temperature T(t) of the roast is uniform throughout. We have T(t)<A=375, T(0)=50, and T(75)=125. Hence

dTdt=k(375T);1375TdT=K dt;ln(375T)=kt+C;375T=Bekt.

Now T(0)=50 implies that B=325, so T(t)=375325ekt. We also know that T=125 when t=75. Substitution of these values in the preceding equation yields

k=175ln(250325)0.0035.

Hence we finally solve the equation

150=375325e(0.0035)t

for t=[ln (225/325)]/(0.0035)105(min), the total cooking time required. Because the roast was placed in the oven at 5:00 p.m., it should be removed at about 6:45 p.m.

Torricelli’s Law

Suppose that a water tank has a hole with area a at its bottom, from which water is leaking. Denote by y(t) the depth of water in the tank at time t, and by V(t) the volume of water in the tank then. It is plausible—and true, under ideal conditions—that the velocity of water exiting through the hole is

(27)v=2gy,

which is the velocity a drop of water would acquire in falling freely from the surface of the water to the hole (see Problem 35 of Section 1.2). One can derive this formula beginning with the assumption that the sum of the kinetic and potential energy of the system remains constant. Under real conditions, taking into account the constriction of a water jet from an orifice, v=c2gy, where c is an empirical constant between 0 and 1 (usually about 0.6 for a small continuous stream of water). For simplicity we take c=1 in the following discussion.

As a consequence of Eq. (27), we have

(28a)dVdt=av=a2gy;

equivalently,

(28b)dVdt=kywherek=a2g.

This is a statement of Torricelli’s law for a draining tank. Let A(y) denote the horizontal cross-sectional area of the tank at height y. Then, applied to a thin horizontal slice of water at height y̲ with area A(y̲) and thickness dy̲, the integral calculus method of cross sections gives

V(y)=0yA(y¯) dy¯.

The fundamental theorem of calculus therefore implies that dV/dy=A(y) and hence that

(29)dVdt=dVdydydt= A(y)dydt.

From Eqs. (28) and (29) we finally obtain

(30)A(y)dydt=a2gy =ky,

an alternative form of Torricelli’s law.

Example 6

Draining tank A hemispherical bowl has top radius 4 ft and at time t=0 is full of water. At that moment a circular hole with diameter 1 in. is opened in the bottom of the tank. How long will it take for all the water to drain from the tank?

Solution

From the right triangle in Fig. 1.4.9, we see that

A(y)=πr2=π[16(4y)2]=π(8yy2).

With g=32 ft/s2, Eq. (30) becomes

π(8yy2)dydt=π(124)2232y;(8y1/2y3/2)dy=172dt;163y3/225y5/2=172t+C.

Now y(0)=4, so

C=163 43/225 45/2=44815.

The tank is empty when y=0, thus when

t=72 448152150 (s);

that is, about 35 min 50 s. So it takes slightly less than 36 min for the tank to drain.

FIGURE 1.4.9.

Draining a hemispherical tank.

Example 7

In the case of an upright cylindrical tank with constant cross-sectional area A, Torricelli’s law in Eq. (30) takes the form

dydt=cy

with c=k/A. With initial condition y(0)=0 we routinely separate variables and integrate to get

y=14c2t2.

However, this formula if taken at face value implies that y>0 if t>0. But if the tank is empty at time t=0 as prescribed by the initial condition y(0)=0, then certainly the tank remains empty thereafter, so y(t)0 for t>0.

To see what is going on here, note that the right-hand side function f(t,y)=cy in our differential equation y=cy does not satisfy the condition that f/y be continuous at (0, 0), so the existence-uniqueness theorem of Section 1.3 does not guarantee uniqueness of a solution near t=0. Indeed, we note the two different but physically meaningful solutions

y1(t)0for all t,andy2(t)={14c2t2for t <0,0for t 0

of the initial value problem y=cy, y(0)=0. The constant solution y1(t)0 corresponds to a tank that always has been and always will be empty, while y2(t) corresponds to a tank draining while t<0 that empties precisely at time t=0 and remains empty thereafter.

Thus this example provides a concrete physical situation described by an initial value problem with non-unique solutions.

1.4 Problems

Find general solutions (implicit if necessary, explicit if convenient) of the differential equations in Problems 1 through 18. Primes denote derivatives with respect to x.

  1. dydx+2xy=0

  2. dydx+2xy2=0

  3. dydx=ysin x

  4. (1+x)dydx=4y

  5. 2xdydx=1y2

  6. dydx=3xy

  7. dydx=(64xy)1/3

  8. dydx=2xsecy

  9. (1x2)dydx=2y

  10. (1+x)2dydx=(1+y)2

  11. y=xy3

  12. yy=x(y2+1)

  13. y3dydx=(y4+1) cos x

  14. dydx=1+x1+y

  15. dydx=(x1)y5x2(2y3y)

  16. (x2+1)(tany)y=x

  17. y=1+x+y+xy (Suggestion: Factor the right-hand side.)

  18. x2y=1x2+y2x2y2

Find explicit particular solutions of the initial value problems in Problems 19 through 28.

  1. dydx=yex, y(0)=2e

  2. dydx=3x2(y2+1),   y(0)=1

  3. 2ydydx=xx216, y(5)=2

  4. dydx=4x3yy, y(1)=3

  5. dydx+1=2y, y(1)=1

  6. dydx=ycotx, y(12π)=12π

  7. xdydxy=2x2y,   y(1)=1

  8. dydx=2xy2+3x2y2, y(1)=1

  9. dydx=6e2xy,y(0)=0

  10. 2xdydx=cos2 y,y(4)=π/4

Problems 29 through 32 explore the connections among general and singular solutions, existence, and uniqueness.

  1. (a) Find a general solution of the differential equation dy/dx=y2. (b) Find a singular solution that is not included in the general solution. (c) Inspect a sketch of typical solution curves to determine the points (a, b) for which the initial value problem y=y2, y(a)=b has a unique solution.

  2. Solve the differential equation (dy/dx)2=4y to verify the general solution curves and singular solution curve that are illustrated in Fig. 1.4.5. Then determine the points (a, b) in the plane for which the initial value problem (y)2=4y, y(a)=b has (a) no solution, (b) infinitely many solutions that are defined for all x, (c) on some neighborhood of the point x=a, only finitely many solutions.

  3. Discuss the difference between the differential equations (dy/dx)2=4y and dy/dx=2y. Do they have the same solution curves? Why or why not? Determine the points (a, b) in the plane for which the initial value problem y=2y, y(a)=b has (a) no solution, (b) a unique solution, (c) infinitely many solutions.

  4. Find a general solution and any singular solutions of the differential equation dy/dx=yy21. Determine the points (a, b) in the plane for which the initial value problem y=yy21, y(a)=b has (a) no solution, (b) a unique solution, (c) infinitely many solutions.

  5. Population growth A certain city had a population of 25,000 in 1960 and a population of 30,000 in 1970. Assume that its population will continue to grow exponentially at a constant rate. What population can its city planners expect in the year 2000?

  6. Population growth In a certain culture of bacteria, the number of bacteria increased sixfold in 10 h. How long did it take for the population to double?

  7. Radiocarbon dating Carbon extracted from an ancient skull contained only one-sixth as much C14 as carbon extracted from present-day bone. How old is the skull?

  8. Radiocarbon dating Carbon taken from a purported relic of the time of Christ contained 4.6×1010 atoms of C14 per gram. Carbon extracted from a present-day specimen of the same substance contained 5.0×1010 atoms of C14 per gram. Compute the approximate age of the relic. What is your opinion as to its authenticity?

  9. Continuously compounded interest Upon the birth of their first child, a couple deposited $5000 in an account that pays 8% interest compounded continuously. The interest payments are allowed to accumulate. How much will the account contain on the child’s eighteenth birthday?

  10. Continuously compounded interest Suppose that you discover in your attic an overdue library book on which your grandfather owed a fine of 30 cents 100 years ago. If an overdue fine grows exponentially at a 5% annual rate compounded continuously, how much would you have to pay if you returned the book today?

  11. Drug elimination Suppose that sodium pentobarbital is used to anesthetize a dog. The dog is anesthetized when its bloodstream contains at least 45 milligrams (mg) of sodium pentobarbitol per kilogram of the dog’s body weight. Suppose also that sodium pentobarbitol is eliminated exponentially from the dog’s bloodstream, with a half-life of 5 h. What single dose should be administered in order to anesthetize a 50-kg dog for 1 h?

  12. Radiometric dating The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable? (Ignore the probable presence of other radioactive isotopes.)

  13. Isotope formation Suppose that a mineral body formed in an ancient cataclysm—perhaps the formation of the earth itself—originally contained the uranium isotope U238 (which has a half-life of 4.51×109 years) but no lead, the end product of the radioactive decay of U238. If today the ratio of U238 atoms to lead atoms in the mineral body is 0.9, when did the cataclysm occur?

  14. Radiometric dating A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (its half-life is about 1.28×109 years) and that one of every nine potassium atom disintegrations yields an argon atom. What is the age of the rock, measured from the time it contained only potassium?

  15. Cooling A pitcher of buttermilk initially at 25°C is to be cooled by setting it on the front porch, where the temperature is 0°C. Suppose that the temperature of the buttermilk has dropped to 15°C after 20 min. When will it be at 5°C?

  16. Solution rate When sugar is dissolved in water, the amount A that remains undissolved after t minutes satisfies the differential equation dA/dt=kA (k>0). If 25% of the sugar dissolves after 1 min, how long does it take for half of the sugar to dissolve?

  17. Underwater light intensity The intensity I of light at a depth of x meters below the surface of a lake satisfies the differential equation dI/dx=(1.4)I. (a) At what depth is the intensity half the intensity I0 at the surface (where x=0)? (b) What is the intensity at a depth of 10 m (as a fraction of I0)? (c) At what depth will the intensity be 1% of that at the surface?

  18. Barometric pressure and altitude The barometric pressure p (in inches of mercury) at an altitude x miles above sea level satisfies the initial value problem dp/dx=(0.2)p, p(0)=29.92. (a) Calculate the barometric pressure at 10,000 ft and again at 30,000 ft. (b) Without prior conditioning, few people can survive when the pressure drops to less than 15 in. of mercury. How high is that?

  19. Spread of rumor A certain piece of dubious information about phenylethylamine in the drinking water began to spread one day in a city with a population of 100,000. Within a week, 10,000 people had heard this rumor. Assume that the rate of increase of the number who have heard the rumor is proportional to the number who have not yet heard it. How long will it be until half the population of the city has heard the rumor?

  20. Isotope formation According to one cosmological theory, when uranium was first generated in the early evolution of the universe following the “big bang,” the isotopes U235 and U238 were produced in equal amounts. Given the half-lives of 4.51×109 years for U238 and 7.10×108 years for U235, calculate the length of time required to reach the present distribution of 137.7 atoms of U238 for each atom of U235.

  21. Cooling A cake is removed from an oven at 210°F and left to cool at room temperature, which is 70°F. After 30 min the temperature of the cake is 140°F. When will it be 100°F?

  22. Pollution increase The amount A(t) of atmospheric pollutants in a certain mountain valley grows naturally and is tripling every 7.5 years.

    1. If the initial amount is 10 pu (pollutant units), write a formula for A(t) giving the amount (in pu) present after t years.

    2. What will be the amount (in pu) of pollutants present in the valley atmosphere after 5 years?

    3. If it will be dangerous to stay in the valley when the amount of pollutants reaches 100 pu, how long will this take?

  23. Radioactive decay An accident at a nuclear power plant has left the surrounding area polluted with radioactive material that decays naturally. The initial amount of radioactive material present is 15 su (safe units), and 5 months later it is still 10 su.

    1. Write a formula giving the amount A(t) of radioactive material (in su) remaining after t months.

    2. What amount of radioactive material will remain after 8 months?

    3. How long—total number of months or fraction thereof—will it be until A=1 su, so it is safe for people to return to the area?

  24. Growth of languages There are now about 3300 different human “language families” in the whole world. Assume that all these are derived from a single original language and that a language family develops into 1.5 language families every 6 thousand years. About how long ago was the single original human language spoken?

  25. Growth of languages Thousands of years ago ancestors of the Native Americans crossed the Bering Strait from Asia and entered the western hemisphere. Since then, they have fanned out across North and South America. The single language that the original Native Americans spoke has since split into many Indian “language families.” Assume (as in Problem 52) that the number of these language families has been multiplied by 1.5 every 6000 years. There are now 150 Native American language families in the western hemisphere. About when did the ancestors of today’s Native Americans arrive?

Torricelli’s Law

Problems 54 through 64 illustrate the application of Torricelli’s law.

  1. A tank is shaped like a vertical cylinder; it initially contains water to a depth of 9 ft, and a bottom plug is removed at time t=0 (hours). After 1 h the depth of the water has dropped to 4 ft. How long does it take for all the water to drain from the tank?

  2. Suppose that the tank of Problem 54 has a radius of 3 ft and that its bottom hole is circular with radius 1 in. How long will it take the water (initially 9 ft deep) to drain completely?

  3. At time t=0 the bottom plug (at the vertex) of a full conical water tank 16 ft high is removed. After 1 h the water in the tank is 9 ft deep. When will the tank be empty?

  4. Suppose that a cylindrical tank initially containing V0 gallons of water drains (through a bottom hole) in T minutes. Use Torricelli’s law to show that the volume of water in the tank after tT minutes is V=V0[1(t/T)]2.

  5. A water tank has the shape obtained by revolving the curve y=x4/3 around the y-axis. A plug at the bottom is removed at 12 noon, when the depth of water in the tank is 12 ft. At 1 p.m. the depth of the water is 6 ft. When will the tank be empty?

  6. A water tank has the shape obtained by revolving the parabola x2=by around the y-axis. The water depth is 4 ft at 12 noon, when a circular plug in the bottom of the tank is removed. At 1 p.m. the depth of the water is 1 ft.(a) Find the depth y(t) of water remaining after t hours.(b) When will the tank be empty?(c) If the initial radius of the top surface of the water is 2 ft, what is the radius of the circular hole in the bottom?

  7. A cylindrical tank with length 5 ft and radius 3 ft is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of water, how long will it take for the liquid to drain completely?

  8. A spherical tank of radius 4 ft is full of water when a circular bottom hole with radius 1 in. is opened. How long will be required for all the water to drain from the tank?

  9. Suppose that an initially full hemispherical water tank of radius 1 m has its flat side as its bottom. It has a bottom hole of radius 1 cm. If this bottom hole is opened at 1 p.m., when will the tank be empty?

  10. Consider the initially full hemispherical water tank of Example 8, except that the radius r of its circular bottom hole is now unknown. At 1 p.m. the bottom hole is opened and at 1:30 p.m. the depth of water in the tank is 2 ft.(a) Use Torricelli’s law in the form dV/dt=(0.6)πr22gy (taking constriction into account) to determine when the tank will be empty. (b) What is the radius of the bottom hole?

  11. A 12 h water clock is to be designed with the dimensions shown in Fig. 1.4.10, shaped like the surface obtained by revolving the curve y=f(x) around the y-axis. What should this curve be, and what should the radius of the circular bottom hole be, in order that the water level will fall at the constant rate of 4 inches per hour (in./h)?

    FIGURE 1.4.10.

    The clepsydra.

  12. Time of death Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of 70°F. At 12 noon the temperature of the body is 80°F and at 1 p.m. it is 75°F. Assume that the temperature of the body at the time of death was 98.6°F and that it has cooled in accord with Newton’s law. What was the time of death?

  13. Snowplow problem Early one morning it began to snow at a constant rate. At 7 a.m. a snowplow set off to clear a road. By 8 a.m. it had traveled 2 miles, but it took two more hours (until 10 a.m.) for the snowplow to go an additional 2 miles.(a) Let t=0 when it began to snow, and let x denote the distance traveled by the snowplow at time t. Assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour, say), show that

    kdxdt=1t

    where k is a constant. (b) What time did it start snowing? (Answer: 6 a.m.)

  14. Snowplow problem A snowplow sets off at 7 a.m. as in Problem 66. Suppose now that by 8 a.m. it had traveled 4 miles and that by 9 a.m. it had moved an additional 3 miles. What time did it start snowing? This is a more difficult snowplow problem because now a transcendental equation must be solved numerically to find the value of k. (Answer: 4:27 a.m.)

  15. Brachistochrone Figure 1.4.11 shows a bead sliding down a frictionless wire from point P to point Q. The brachistochrone problem asks what shape the wire should be in order to minimize the bead’s time of descent from P to Q. In June of 1696, John Bernoulli proposed this problem as a public challenge, with a 6-month deadline (later extended to Easter 1697 at George Leibniz’s request). Isaac Newton, then retired from academic life and serving as Warden of the Mint in London, received Bernoulli’s challenge on January 29, 1697. The very next day he communicated his own solution—the curve of minimal descent time is an arc of an inverted cycloid—to the Royal Society of London. For a modern derivation of this result, suppose the bead starts from rest at the origin P and let y=y(x) be the equation of the desired curve in a coordinate system with the y-axis pointing downward. Then a mechanical analogue of Snell’s law in optics implies that

    (i)sinαv= constant,

    where α denotes the angle of deflection (from the vertical) of the tangent line to the curve—so cotα=y(x) (why?)—and v=2gy is the bead’s velocity when it has descended a distance y vertically (from KE=12mv2=mgy=PE).

    FIGURE 1.4.11.

    A bead sliding down a wire—the brachistochrone problem.

    1. First derive from Eq. (i) the differential equation

      (ii)dydx=2ayy,

      where a is an appropriate positive constant.

    2. Substitute y=2asin2 t, dy=4asin tcos t dt in (ii) to derive the solution

      (iii)x=a(2tsin 2t),    y=a(1cos 2t)

      for which t=y=0 when x=0. Finally, the substitution of θ=2t in (iii) yields the standard parametric equations x=a(θsin θ), y=a(1cos θ) of the cycloid that is generated by a point on the rim of a circular wheel of radius a as it rolls along the x-axis. [See Example 5 in Section 9.4 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008.]

  16. Hanging cable Suppose a uniform flexible cable is suspended between two points (±L,H) at equal heights located symmetrically on either side of the x-axis (Fig. 1.4.12). Principles of physics can be used to show that the shape y=y(x) of the hanging cable satisfies the differential equation

    ad2ydx2=1+(dydx)2,

    where the constant a=T/ρ is the ratio of the cable’s tension T at its lowest point x=0 (where y(0)=0) and its (constant) linear density ρ. If we substitute v=dy/dx, dv/dx=d2y/dx2 in this second-order differential equation, we get the first-order equation

    advdx=1+v2.

    Solve this differential equation for y(x)=v(x)=sinh(x/a). Then integrate to get the shape function

    y(x)= a cosh(xa) +C

    of the hanging cable. This curve is called a catenary, from the Latin word for chain.

    FIGURE 1.4.12.

    The catenary.

1.4 Application The Logistic Equation

As in Eq. (7) of this section, the solution of a separable differential equation reduces to the evaluation of two indefinite integrals. It is tempting to use a symbolic algebra system for this purpose. We illustrate this approach using the logistic differential equation

(1)dxdt=axbx2

that models a population x(t) with births (per unit time) proportional to x and deaths proportional to x2. Here we concentrate on the solution of Eq. (1) and defer discussion of population applications to Section 2.1.

If a=0.01 and b=0.0001, for instance, Eq. (1) is

(2)dxdt=(0.01)x(0.0001)x2=x10000(100x).

Separation of variables leads to

(3)1x(100x)dx=110000dt=t10000+C.

We can evaluate the integral on the left by using the Maple command

FIGURE 1.4.13.

TI-89 screen showing the integral in Eq. (3).

int(1/(x∗(100 - x)), x);

the Mathematica command

Integrate[ 1/(x∗(100 - x)), x ]

or the Matlab command

syms x; int(1/(x∗(100 - x) ) )

Alternatively, we could use the freely available Wolfram|Alpha system (www.wolframalpha.com); the query

integrate 1/(x*(100 - x) )

produces the output shown in Fig. 1.4.14.

FIGURE 1.4.14.

Wolfram|Alpha display showing the integral in Eq. (3). Screenshot of Wolfram|Alpha output. Used by permission of WolframAlpha LLC.

Any computer algebra system gives a result of the form

(4)1100ln x1100ln(x100)=t10000+C

equivalent to the graphing calculator result shown in Fig. 1.4.13.

You can now apply the initial condition x(0)=x0, combine logarithms, and finally exponentiate to solve Eq. (4) for the particular solution

(5)x(t)=100x0et/100100x0+x0et/100

of Eq. (2). The slope field and solution curves shown in Fig. 1.4.15 suggest that, whatever is the initial value x0, the solution x(t) approaches 100 as t+. Can you use Eq. (5) to verify this conjecture?

FIGURE 1.4.15.

Slope field and solution curves for x=(0.01)x(0.0001)x2.

Investigation: For your own personal logistic equation, take a=m/n and b=1/n in Eq. (1), with m and n being the largest two distinct digits (in either order) in your student ID number.

  1. First generate a slope field for your differential equation and include a sufficient number of solution curves that you can see what happens to the population as t+. State your inference plainly.

  2. Next use a computer algebra system to solve the differential equation symbolically; then use the symbolic solution to find the limit of x(t) as t+. Was your graphically based inference correct?

  3. Finally, state and solve a numerical problem using the symbolic solution. For instance, how long does it take x to grow from a selected initial value x0 to a given target value x1?

1.5 Linear First-Order Equations

We turn now to another important method for solving first-order differential equations that rests upon the idea of “integrating both sides.” In Section 1.4 we saw that the first step in solving a separable differential equation is to multiply and/or divide both sides of the equation by whatever is required in order to separate the variables. For instance, to solve the equation

(1)dydx=2xy(y>0),

we divide both sides by y (and, so to speak, multiply by the differential dx) to get

dyy=2x dx.

Integrating both sides then gives the general solution ln y=x2+C.

There is another way to approach the differential equation in (1), however, which—while leading to the same general solution—opens the door not only to the solution method discussed in this section, but to other methods of solving differential equations as well. What is common to all these methods is the idea that if a given equation is difficult to solve, then perhaps multiplying both sides of the equation by a suitably chosen function of x and/or y may result in an equivalent equation that can be solved more easily. Thus, in Eq. (1), rather than divide both sides by y, we could instead multiply both sides by the factor 1/y. (Of course algebraically these two are the same, but we are highlighting the fact that often the crucial first step in solving a differential equation is to multiply both of its sides by the “right” function.) Applying this to Eq. (1) (while leaving dx in place) gives

(2)1y.dydx=2x.

The significance of Eq. (2) is that, unlike Eq. (1), both sides are recognizable as a derivative. By the chain rule, the left-hand side of Eq. (2) can be written as

1y·dydx=Dx(ln y),

and of course the right hand side of Eq. (2) is Dx(x2). Thus each side of Eq. (2) can be viewed as a derivative with respect to x:

Dx(ln y)=Dx(x2).

Integrating both sides with respect to x gives the same general solution ln y=x2+C that we found before.

We were able to solve the differential equation in Eq. (1), then, by first multiplying both of its sides by a factor—known as an integrating factor—chosen so that both sides of the resulting equation could be recognized as a derivative. Solving the equation then becomes simply a matter of integrating both sides. More generally, an integrating factor for a differential equation is a function ρ(x,y) such that multiplication of each side of the differential equation by ρ(x,y) yields an equation in which each side is recognizable as a derivative. In some cases integrating factors involve both of the variables x and y; however, our second solution of Eq. (1) was based on the integrating factor ρ(y)=1/y, which depends only on y. Our goal in this section is to show how integrating factors can be used to solve a broad and important category of first-order differential equations.

A linear first-order equation is a differential equation of the form

(3)dydx+P(x)y=Q(x).

We assume that the coefficient functions P(x) and Q(x) are continuous on some interval on the x-axis. (Can you see that the differential equation in Eq. (1), in addition to being separable, is also linear? Is every separable equation also linear?) Assuming that the necessary antiderivatives can be found, the general linear equation in (3) can always be solved by multiplying by the integrating factor

(4)ρ(x)=eP(x)dx.

The result is

(5)eP(x)dxdydx+P(x)eP(x)dxy=Q(x)eP(x)dx.

Because

Dx[P(x)dx] =P(x),

the left-hand side is the derivative of the product y(x)·eP(x)dx, so Eq. (5) is equivalent to

Dx[y(x).eP(x)dx] =Q(x)eP(x)dx.

Integration of both sides of this equation gives

y(x)eP(x)dx=(Q(x)eP(x)dx) dx+C.

Finally, solving for y, we obtain the general solution of the linear first-order equation in (3):

(6)y(x)=eP(x)dx=[(Q(x)eP(x)dx) dx+C.]

This formula should not be memorized. In a specific problem it generally is simpler to use the method by which we developed the formula. That is, in order to solve an equation that can be written in the form in Eq. (3) with the coefficient functions P(x) and Q(x) displayed explicitly, you should attempt to carry out the following steps.

METHOD: SOLUTION OF LINEAR FIRST-ORDER EQUATIONS

  1. Begin by calculating the integrating factor ρ(x)=eP(x)dx.

  2. Then multiply both sides of the differential equation by ρ(x).

  3. Next, recognize the left-hand side of the resulting equation as the derivative of a product:

    Dx [ρ(x)y(x)]=ρ(x)Q(x).
  4. Finally, integrate this equation,

    ρ(x)y(x)=ρ(x) Q(x) dx+C,

    then solve for y to obtain the general solution of the original differential equation.

Remark 1

Given an initial condition y(x0)=y0, you can (as usual) substitute x=x0 and y=y0 into the general solution and solve for the value of C yielding the particular solution that satisfies this initial condition.

Remark 2

You need not supply explicitly a constant of integration when you find the integrating factor ρ(x). For if we replace

P(x)dxwithP(x)dx+K

in Eq. (4), the result is

ρ(x)=eK+P(x)dx=eKeP(x)dx.

But the constant factor eK does not affect materially the result of multiplying both sides of the differential equation in (3) by ρ(x), so we might as well take K=0. You may therefore choose for P(x)dx any convenient antiderivative of P(x), without bothering to add a constant of integration.

Example 1

Solve the initial value problem

dydxy=118ex/3,y(0)= 1.

Solution

Here we have P(x)1 and Q(x)=118ex/3, so the integrating factor is

ρ(x)=e(1)dx=ex.

Multiplication of both sides of the given equation by ex yields

(7)exdydxexy=118e4x/3,

which we recognize as

ddx(exy)=118e4x/3.

Hence integration with respect to x gives

exy=118e4x/3dx=3332e4x/3+C,

and multiplication by ex gives the general solution

(8)y(x)=Cex3332ex/3.

Substitution of x=0 and y=1 now gives C=132, so the desired particular solution is

y(x)=132ex3332ex/3=132(ex33ex/3).

Remark

Figure 1.5.1 shows a slope field and typical solution curves for Eq. (7), including the one passing through the point (0,1). Note that some solutions grow rapidly in the positive direction as x increases, while others grow rapidly in the negative direction. The behavior of a given solution curve is determined by its initial condition y(0)=y0. The two types of behavior are separated by the particular solution y(x)=3332ex/3 for which C=0 in Eq. (8), so y0=3332 for the solution curve that is dashed in Fig. 1.5.1. If y0>3332, then C>0 in Eq. (8), so the term ex eventually dominates the behavior of y(x), and hence y(x)+ as x+. But if y0<3332, then C<0, so both terms in y(x) are negative and therefore y(x) as x+. Thus the initial condition y0=3332 is critical in the sense that solutions that start above 3332 on the y-axis grow in the positive direction, while solutions that start lower than 3332 grow in the negative direction as x+. The interpretation of a mathematical model often hinges on finding such a critical condition that separates one kind of behavior of a solution from a different kind of behavior.

FIGURE 1.5.1.

Slope field and solution curves for y=y+118ex/3.

Example 2

Find a general solution of

(9)(x2+1)dydx+3xy=6x.

Solution

After division of both sides of the equation by x2+1, we recognize the result

dydx+3xx2+1 y =6xx2+1

as a first-order linear equation with P(x)=3x/(x2+1) and Q(x)=6x/(x2+1). Multiplication by

ρ(x)=exp (3xx2+1dx)=exp(32ln(x2+1))=(x2+1)3/2

yields

(x2+1)3/2dydx+3x(x2+1)1/2y=6x(x2+1)1/2,

and thus

Dx[(x2+1)3/2y]=6x(x2+1)1/2.

Integration then yields

(x2+1)3/2y=6x(x2+1)1/2dx=2(x2+1)3/2+C.

Multiplication of both sides by (x2+1)3/2 gives the general solution

(10)y(x)=2+C(x2+1)3/2.

Remark

Figure 1.5.2 shows a slope field and typical solution curves for Eq. (9). Note that, as x+, all other solution curves approach the constant solution curve y(x)2 that corresponds to C=0 in Eq. (10). This constant solution can be described as an equilibrium solution of the differential equation, because y(0)=2 implies that y(x)=2 for all x (and thus the value of the solution remains forever where it starts). More generally, the word “equilibrium” connotes “unchanging,” so by an equilibrium solution of a differential equation is meant a constant solution y(x)c, for which it follows that y(x)0. Note that substitution of y=0 in the differential equation (9) yields 3xy=6x, so it follows that y=2 if x0. Hence we see that y(x)2 is the only equilibrium solution of this differential equation, as seems visually obvious in Fig. 1.5.2.

FIGURE 1.5.2.

Slope field and solution curves for the differential equation in Eq. (9).

A Closer Look at the Method

The preceding derivation of the solution in Eq. (6) of the linear first-order equation y+Py=Q bears closer examination. Suppose that the coefficient functions P(x) and Q(x) are continuous on the (possibly unbounded) open interval I. Then the antiderivatives

P(x)dx and(Q(x)eP(x)dx) dx

exist on I. Our derivation of Eq. (6) shows that if y=y(x) is a solution of Eq. (3) on I, then y(x) is given by the formula in Eq. (6) for some choice of the constant C. Conversely, you may verify by direct substitution (Problem 31) that the function y(x) given in Eq. (6) satisfies Eq. (3). Finally, given a point x0 of I and any number y0, there is—as previously noted—a unique value of C such that y(x0)=y0. Consequently, we have proved the following existence-uniqueness theorem.

Remark 1

Theorem 1 gives a solution on the entire interval I for a linear differential equation, in contrast with Theorem 1 of Section 1.3, which guarantees only a solution on a possibly smaller interval.

Remark 2

Theorem 1 tells us that every solution of Eq. (3) is included in the general solution given in Eq. (6). Thus a linear first-order differential equation has no singular solutions.

Remark 3

The appropriate value of the constant C in Eq. (6)—as needed to solve the initial value problem in Eq. (11)—can be selected “automatically” by writing

(12)ρ(x)=exp(x0xP(t)dt),y(x)=1ρ(x)[y0+x0xρ(t)Q(t)dt].

The indicated limits x0 and x effect a choice of indefinite integrals in Eq. (6) that guarantees in advance that ρ(x0)=1 and that y(x0)=y0 (as you can verify directly by substituting x=x0 in Eq. (12)).

Example 3

Solve the initial value problem

(13)x2dydx+xy=sin x,y(1)=y0.

Solution

Division by x2 gives the linear first-order equation

dydx + 1x y = sin xx2

with P(x)=1/x and Q(x)=(sin x)/x2. With x0=1 the integrating factor in (12) is

ρ(x)=exp(1x1tdt)=exp(ln x)=x,

so the desired particular solution is given by

(14)y(x)=1x[y0+1xsin tt dt].

In accord with Theorem 1, this solution is defined on the whole positive x-axis.

Comment

In general, an integral such as the one in Eq. (14) would (for given x) need to be approximated numerically—using Simpson’s rule, for instance—to find the value y(x) of the solution at x. In this case, however, we have the sine integral function

Si(x)=0xsin tt dt,

which appears with sufficient frequency in applications that its values have been tabulated. A good set of tables of special functions is Abramowitz and Stegun, Handbook of Mathematical Functions (New York: Dover, 1965). Then the particular solution in Eq. (14) reduces to

(15)y(x)=1x[y0+0xsin tt dt01sin tt dt]=1x[y0+Si(x)Si(1)].

The sine integral function is available in most scientific computing systems and can be used to plot typical solution curves defined by Eq. (15). Figure 1.5.3 shows a selection of solution curves with initial values y(1)=y0 ranging from y0=3 to y0=3. It appears that on each solution curve, y(x)0 as x+, and this is in fact true because the sine integral function is bounded.

FIGURE 1.5.3.

Typical solution curves defined by Eq. (15).

In the sequel we will see that it is the exception—rather than the rule—when a solution of a differential equation can be expressed in terms of elementary functions. We will study various devices for obtaining good approximations to the values of the nonelementary functions we encounter. In Chapter 2 we will discuss numerical integration of differential equations in some detail.

Mixture Problems

As a first application of linear first-order equations, we consider a tank containing a solution—a mixture of solute and solvent—such as salt dissolved in water. There is both inflow and outflow, and we want to compute the amount x(t) of solute in the tank at time t, given the amount x(0)=x0 at time t=0. Suppose that solution with a concentration of ci grams of solute per liter of solution flows into the tank at the constant rate of ri liters per second, and that the solution in the tank—kept thoroughly mixed by stirring—flows out at the constant rate of r0 liters per second.

To set up a differential equation for x(t), we estimate the change Δx in x during the brief time interval [t,t+Δt]. The amount of solute that flows into the tank during Δt seconds is riciΔt grams. To check this, note how the cancellation of dimensions checks our computations:

(riliterssecond)(cigramsliter) (Δt seconds)

yields a quantity measured in grams.

FIGURE 1.5.4.

The single-tank mixture problem.

The amount of solute that flows out of the tank during the same time interval depends on the concentration c0(t) of solute in the solution at time t. But as noted in Fig. 1.5.4, c0(t)=x(t)/V(t), where V(t) denotes the volume (not constant unless ri=r0) of solution in the tank at time t. Then

Δx={grams input}{grams output}rici Δtr0c0Δt.

We now divide by Δt:

ΔxΔtricir0c0.

Finally, we take the limit as Δt0; if all the functions involved are continuous and x(t) is differentiable, then the error in this approximation also approaches zero, and we obtain the differential equation

(16)dxdtricir0c0,

in which ri, ci, and r0 are constants, but c0 denotes the variable concentration

(17)c0(t)=x(t)V(t)

of solute in the tank at time t. Thus the amount x(t) of solute in the tank satisfies the differential equation

(18)dxdt=ricir0Vx.

If V0=V(0), then V(t)=V0+(rir0)t, so Eq. (18) is a linear first-order differential equation for the amount x(t) of solute in the tank at time t.

Important

Equation (18) need not be committed to memory. It is the process we used to obtain that equation—examination of the behavior of the system over a short time interval [t,t+Δt]—that you should strive to understand, because it is a very useful tool for obtaining all sorts of differential equations.

Remark

It was convenient for us to use g/L mass/volume units in deriving Eq. (18). But any other consistent system of units can be used to measure amounts of solute and volumes of solution. In the following example we measure both in cubic kilometers.

Example 4

Mixture problem Assume that Lake Erie has a volume of 480 km3 and that its rate of inflow (from Lake Huron) and outflow (to Lake Ontario) are both 350 km3 per year. Suppose that at the time t=0 (years), the pollutant concentration of Lake Erie—caused by past industrial pollution that has now been ordered to cease—is five times that of Lake Huron. If the outflow henceforth is perfectly mixed lake water, how long will it take to reduce the pollution concentration in Lake Erie to twice that of Lake Huron?

Solution

Here we have

V=480 (km3),ri=r0=r=350 (km3/yr),ci=c (the pollutant concentration of Lake Huron), andx0=x(0)=5cV,

and the question is this: When is x(t)=2cV? With this notation, Eq. (18) is the separable equation

(19)dxdt=rcrVx,

which we rewrite in the linear first-order form

(20)dxdt+px=q

with constant coefficients p=r/V, q=rc, and integrating factor ρ=ept. You can either solve this equation directly or apply the formula in (12). The latter gives

(21)x(t)=ept[x0+0tqeptdt]=ept[x0+qp(ept1)]=ert/V[5cV+rcr/V(ert/V1)];x(t)=cV+4cVert/V.

To find when x(t)=2cV, we therefore need only solve the equation

cV+4cVert/V=2cVfort=Vrln 4=480350ln 41.901 (years).

Example 5

Mixture problem A 120-gallon (gal) tank initially contains 90 lb of salt dissolved in 90 gal of water. Brine containing 2 lb/gal of salt flows into the tank at the rate of 4 gal/min, and the well-stirred mixture flows out of the tank at the rate of 3 gal/min. How much salt does the tank contain when it is full?

Solution

The interesting feature of this example is that, due to the differing rates of inflow and outflow, the volume of brine in the tank increases steadily with V(t)=90+t gallons. The change Δx in the amount x of salt in the tank from time t to time t+Δt (minutes) is given by

Δ x (4) (2)Δt  3(x90+t)Δt,

so our differential equation is

dxdt+390+tx=8.

An integrating factor is

ρ(x)=exp(390+tdt)=e3 ln(90+t)=(90+t)3,

which gives

Dt[(90+t)3x]=8(90+t)3;(90+t)3x=2(90+t)4+C.

Substitution of x(0)=90 gives C=(90)4, so the amount of salt in the tank at time t is

x(t)=2(90+t)904(90+t)3.

The tank is full after 30 min, and when t=30, we have

x(30)=2(90+30)9041203202(lb)

of salt in the tank.

1.5 Problems

Find general solutions of the differential equations in Problems 1 through 25. If an initial condition is given, find the corresponding particular solution. Throughout, primes denote derivatives with respect to x.

  1. y+y=2, y(0)=0

  2. y2y=3e2x, y(0)=0

  3. y+3y=2xe3x

  4. y2xy=ex2

  5. xy+2y=3x, y(1)=5

  6. xy+5y=7x2, y(2)=5

  7. 2xy+y=10x

  8. 3xy+y=12x

  9. xyy=x,y(1)=7

  10. 2xy3y=9x3

  11. xy+y=3xy, y(1)=0

  12. xy+3y=2x5, y(2)=1

  13. y+y=ex, y(0)=1

  14. xy3y=x3, y(1)=10

  15. y+2xy=x, y(0)=2

  16. y=(1y) cos x, y(π)=2

  17. (1+x)y+y=cos x, y(0)=1

  18. xy=2y+x3 cos x

  19. y+ycotx=cos x

  20. y=1+x+y+xy, y(0)=0

  21. xy=3y+x4 cos x, y(2π)=0

  22. y=2xy+3x2exp(x2), y(0)=5

  23. xy+(2x3)y=4x4

  24. (x2+4)y+3xy=x, y(0)=1

  25. (x2+1)dydx+3x3y=6xexp(32x2),y(0)=1

Solve the differential equations in Problems 26 through 28 by regarding y as the independent variable rather than x.

  1. (14xy2)dydx=y3

  2. (x+yey)dydx=1

  3. (1+2xy)dydx=1+y2

  4. Express the general solution of dy/dx=1+2xy in terms of the error function

    erf(x)=2π 0xet2dt.
  5. Express the solution of the initial value problem

    2xdydx=y+2x cos x,y(1) = 0

    as an integral as in Example 3 of this section.

Problems 31 and 32 illustrate—for the special case of first-order linear equations-techniques that will be important when we study higher—order linear equations in Chapter 3.

  1. (a) Show that

    yc(x)=CeP(x) dx

    is a general solution of dy/dx+P(x)y=0. (b) Show that

    yp(x)=eP(x) dx[(Q(x)eP(x) dx)dx]

    is a particular solution of dy/dx+P(x)y=Q(x). (c) Suppose that yc(x) is any general solution of dy/dx+P(x)y=0 and that yp(x) is any particular solution of dy/dx+P(x)y=Q(x). Show that y(x)=yc(x)+yp(x) is a general solution of dy/dx+P(x)y=Q(x).

  2. (a) Find constants A and B such that yp(x)=A sin x+B cos x is a solution of dy/dx+y=2 sin x. (b) Use the result of part (a) and the method of Problem 31 to find the general solution of dy/dx+y=2 sin x. (c) Solve the initial value problem dy/dx+y=2 sin x, y(0)=1.

Mixture Problems

Problems 33 through 37 illustrate the application of linear first-order differential equations to mixture problems.

  1. A tank contains 1000 liters (L) of a solution consisting of 100 kg of salt dissolved in water. Pure water is pumped into the tank at the rate of 5 L/s, and the mixture—kept uniform by stirring—is pumped out at the same rate. How long will it be until only 10 kg of salt remains in the tank?

  2. Consider a reservoir with a volume of 8 billion cubic feet (ft3) and an initial pollutant concentration of 0.25%. There is a daily inflow of 500 million ft3 of water with a pollutant concentration of 0.05% and an equal daily outflow of the well-mixed water in the reservoir. How long will it take to reduce the pollutant concentration in the reservoir to 0.10%?

  3. Rework Example 4 for the case of Lake Ontario, which empties into the St. Lawrence River and receives inflow from Lake Erie (via the Niagara River). The only differences are that this lake has a volume of 1640 km3 and an inflow-outflow rate of 410 km3/ year.

  4. A tank initially contains 60 gal of pure water. Brine containing 1 lb of salt per gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at 3 gal/min; thus the tank is empty after exactly 1 h. (a) Find the amount of salt in the tank after t minutes. (b) What is the maximum amount of salt ever in the tank?

  5. A 400-gal tank initially contains 100 gal of brine containing 50 lb of salt. Brine containing 1 lb of salt per gallon enters the tank at the rate of 5 gal/s, and the well-mixed brine in the tank flows out at the rate of 3 gal/s. How much salt will the tank contain when it is full of brine?

  6. Two tanks Consider the cascade of two tanks shown in Fig. 1.5.5, with V1=100 (gal) and V2=200 (gal) the volumes of brine in the two tanks. Each tank also initially contains 50 lb of salt. The three flow rates indicated in the figure are each 5 gal/min, with pure water flowing into tank 1. (a) Find the amount x(t) of salt in tank 1 at time t.(b) Suppose that y(t) is the amount of salt in tank 2 at time t. Show first that

    dydt=5x1005y200,

    and then solve for y(t), using the function x(t) found in part (a).(c) Finally, find the maximum amount of salt ever in tank 2.

    FIGURE 1.5.5.

    A cascade of two tanks.

  7. Two tanks Suppose that in the cascade shown in Fig. 1.5.5, tank 1 initially contains 100 gal of pure ethanol and tank 2 initially contains 100 gal of pure water. Pure water flows into tank 1 at 10 gal/min, and the other two flow rates are also 10 gal/min.(a) Find the amounts x(t) and y(t) of ethanol in the two tanks at time t0. (b) Find the maximum amount of ethanol ever in tank 2.

  8. Multiple tanks A multiple cascade is shown in Fig. 1.5.6. At time t=0, tank 0 contains 1 gal of ethanol and 1 gal of water; all the remaining tanks contain 2 gal of pure water each. Pure water is pumped into tank 0 at 1 gal/min, and the varying mixture in each tank is pumped into the one below it at the same rate. Assume, as usual, that the mixtures are kept perfectly uniform by stirring. Let xn(t) denote the amount of ethanol in tank n at time t.

    FIGURE 1.5.6.

    A multiple cascade.

    (a) Show that x0(t)=et/2. (b) Show by induction on n that

    xn(t)=tnet/2n!2nfor n0.

    (c) Show that the maximum value of xn(t) for n>0 is Mn=xn(2n)=nnen/n!. (d) Conclude from Stirling’s approximation n!nnen2πn that Mn(2πn)1/2.

  9. Retirement savings A 30-year-old woman accepts an engineering position with a starting salary of $30,000 per year. Her salary S(t) increases exponentially, with S(t)=30et/20 thousand dollars after t years. Meanwhile, 12% of her salary is deposited continuously in a retirement account, which accumulates interest at a continuous annual rate of 6%.(a) Estimate ΔA in terms of Δt to derive the differential equation satisfied by the amount A(t) in her retirement account after t years.(b) Compute A(40), the amount available for her retirement at age 70.

  10. Falling hailstone Suppose that a falling hailstone with density δ=1 starts from rest with negligible radius r=0. Thereafter its radius is r=kt (k is a constant) as it grows by accretion during its fall. Use Newton#x2019;s second law—according to which the net force F acting on a possibly variable mass m equals the time rate of change dp/dt of its momentum p=mv—to set up and solve the initial value problem

    ddt(mv)=mg,v(0)=0, 

    where m is the variable mass of the hailstone, v=dy/dt is its velocity, and the positive y-axis points downward. Then show that dv/dt=g/4. Thus the hailstone falls as though it were under one-fourth the influence of gravity.

  11. Figure 1.5.7 shows a slope field and typical solution curves for the equation y=xy. (a) Show that every solution curve approaches the straight line y=x1 as x+. (b) For each of the five values y1=3.998, 3.999, 4.000, 4.001, and 4.002, determine the initial value y0 (accurate to four decimal places) such that y(5)=y1 for the solution satisfying the initial condition y(5)=y0.

    FIGURE 1.5.7.

    Slope field and solution curves for y=xy.

  12. Figure 1.5.8 shows a slope field and typical solution curves for the equation y=x+y. (a) Show that every solution curve approaches the straight line y=x1 as x. (b) For each of the five values y1=10, 5, 0, 5, and 10, determine the initial value y0 (accurate to five decimal places) such that y(5)=y1 for the solution satisfying the initial condition y(5)=y0.

    FIGURE 1.5.8.

    Slope field and solution curves for y=x+y.

Polluted Reservoir

Problems 45 and 46 deal with a shallow reservoir that has a one-square-kilometer water surface and an average water depth of 2 meters. Initially it is filled with fresh water, but at time t=0 water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month. The well-mixed water in the reservoir flows out at the same rate. Your first task is to find the amount x(t) of pollutant (in millions of liters) in the reservoir after t months.

  1. The incoming water has a pollutant concentration of c(t)=10 liters per cubic meter (L/m3). Verify that the graph of x(t) resembles the steadily rising curve in Fig. 1.5.9, which approaches asymptotically the graph of the equilibrium solution x(t)20 that corresponds to the reservoir’s long-term pollutant content. How long does it take the pollutant concentration in the reservoir to reach 10 L/m3?

  2. The incoming water has pollutant concentration c(t)=10(1+cos t) L/m3 that varies between 0 and 20, with an average concentration of 10 L/m3 and a period of oscillation of slightly over 614 months. Does it seem predictable that the lake’s pollutant content should ultimately oscillate periodically about an average level of 20 million liters? Verify that the graph of x(t) does, indeed, resemble the oscillatory curve shown in Fig. 1.5.9. How long does it take the pollutant concentration in the reservoir to reach 10 L/m3?

    FIGURE 1.5.9.

    Graphs of solutions in Problems 45 and 46.

1.5 Application Indoor Temperature Oscillations

For an interesting applied problem that involves the solution of a linear differential equation, consider indoor temperature oscillations that are driven by outdoor temperature oscillations of the form

(1)A(t)=a0+a1cosωt+b1sinωt.

If ω=π/12, then these oscillations have a period of 24 hours (so that the cycle of outdoor temperatures repeats itself daily) and Eq. (1) provides a realistic model for the temperature outside a house on a day when no change in the overall day-to-day weather pattern is occurring. For instance, for a typical July day in Athens, Georgia with a minimum temperature of 70°F when t=4 (4 a.m.) and a maximum of 90°F when t=16 (4 p.m.), we would take

(2)A(t)=80 10 cos ωt(t 4)=805 cos ωt5 3 sin ωt.

We derived Eq. (2) by using the identity cos(αβ)=cos αcos β+sin αsin β to get a0=80, a1=5, and b1=53 in Eq. (1).

If we write Newton’s law of cooling (Eq. (3) of Section 1.1) for the corresponding indoor temperature u(t) at time t, but with the outside temperature A(t) given by Eq. (1) instead of a constant ambient temperature A, we get the linear first-order differential equation

dudt=k(uA(t));

that is,

(3)dudt+ku=k(a0+a1cosωt+b1sinωt)

with coefficient functions P(t)k and Q(t)=kA(t). Typical values of the proportionality constant k range from 0.2 to 0.5 (although k might be greater than 0.5 for a poorly insulated building with open windows, or less than 0.2 for a well-insulated building with tightly sealed windows).

Scenario: Suppose that our air conditioner fails at time t0=0 one midnight, and we cannot afford to have it repaired until payday at the end of the month. We therefore want to investigate the resulting indoor temperatures that we must endure for the next several days.

Begin your investigation by solving Eq. (3) with the initial condition u(0)=u0 (the indoor temperature at the time of the failure of the air conditioner). You may want to use the integral formulas in 49 and 50 of the endpapers, or possibly a computer algebra system. You should get the solution

(4)u(t)=a0+c0ekt+c1cosωt+d1sinωt,

where

c0=u0a0k2a1kωb1k2+ω2,c1=k2a1kωb1k2+ω2,d1=kωa1+k2b1k2+ω2

with ω=π/12.

With a0=80, a1=5, b1=53 (as in Eq. (2)), ω=π/12, and k=0.2 (for instance), this solution reduces (approximately) to

(5)u(t)=80+et/5(u082.3351)+(2.3351)cosπt12(5.6036)sinπt12.

Observe first that the “damped” exponential term in Eq. (5) approaches zero as t+, leaving the long-term “steady periodic” solution

(6)usp(t)=80+(2.3351)cosπt12(5.6036)sinπt12.

Consequently, the long-term indoor temperatures oscillate every 24 hours around the same average temperature 80°F as the average outdoor temperature.

Figure 1.5.10 shows a number of solution curves corresponding to possible initial temperatures u0 ranging from 65°F to 95°F. Observe that—whatever the initial temperature—the indoor temperature “settles down” within about 18 hours to a periodic daily oscillation. But the amplitude of temperature variation is less indoors than outdoors. Indeed, using the trigonometric identity mentioned earlier, Eq. (6) can be rewritten (verify this!) as

FIGURE 1.5.10.

Solution curves given by Eq. (5) with u0=65,68,71,,92,95.

(7)u(t)=80(6.0707) cos(πt121.9656)=80(6.0707) cosπ2(t7.5082).

Do you see that this implies that the indoor temperature varies between a minimum of about 74°F and a maximum of about 86°F?

Finally, comparison of Eqs. (2) and (7) indicates that the indoor temperature lags behind the outdoor temperature by about 7.508243.5 hours, as illustrated in Fig. 1.5.11. Thus the temperature inside the house continues to rise until about 7:30 p.m. each evening, so the hottest part of the day inside is early evening rather than late afternoon (as outside).

FIGURE 1.5.11.

Comparison of indoor and outdoor temperature oscillations.

For a personal problem to investigate, carry out a similar analysis using average July daily maximum/minimum figures for your own locale and a value of k appropriate to your own home. You might also consider a winter day instead of a summer day. (What is the winter-summer difference for the indoor temperature problem?) You may wish to explore the use of available technology both to solve the differential equation and to graph its solution for the indoor temperature in comparison with the outdoor temperature.

1.6 Substitution Methods and Exact Equations

The first-order differential equations we have solved in the previous sections have all been either separable or linear. But many applications involve differential equations that are neither separable nor linear. In this section we illustrate (mainly with examples) substitution methods that sometimes can be used to transform a given differential equation into one that we already know how to solve.

For instance, the differential equation

(1)dydx=f (x,y) ,

with dependent variable y and independent variable x, may contain a conspicuous combination

(2)v=α(x,y)

of x and y that suggests itself as a new independent variable v. Thus the differential equation

dydx=(x+y+3)2

practically demands the substitution v=x+y+3 of the form in Eq. (2).

If the substitution relation in Eq. (2) can be solved for

(3)y=β(x,v),

then application of the chain rule—regarding v as an (unknown) function of x—yields

(4)dydx=βxdxdx+βvdvdx=βx+βvdvdx,

where the partial derivatives β/x=βx(x,v) and β/v=βv(x,v) are known functions of x and v. If we substitute the right-hand side in (4) for dy/dx in Eq. (1) and then solve for dv/dx, the result is a new differential equation of the form

(5)dvdx=g(x,v)

with new dependent variable v. If this new equation is either separable or linear, then we can apply the methods of preceding sections to solve it.

If v=v(x) is a solution of Eq. (5), then y=β(x,v(x)) will be a solution of the original Eq. (1). The trick is to select a substitution such that the transformed Eq. (5) is one we can solve. Even when possible, this is not always easy; it may require a fair amount of ingenuity or trial and error.

Example 1

Solve the differential equation

dydx=(x+y+3)2.

Solution

As indicated earlier, let’s try the substitution

v=x+y+3;that is,y=vx3.

Then

dydx=dvdx1,

so the transformed equation is

dvdx=1+v2.

This is a separable equation, and we have no difficulty in obtaining its solution

x=dv1+v2=tan1v+C.

So v=tan(xC). Because v=x+y+3, the general solution of the original equation dy/dx=(x+y+3)2 is x+y+3=tan(xC); that is,

y(x)=tan(x C) x 3.

Remark

Figure 1.6.1 shows a slope field and typical solution curves for the differential equation of Example 1. We see that, although the function f(x,y)=(x+y+3)2 is continuously differentiable for all x and y, each solution is continuous only on a bounded interval. In particular, because the tangent function is continuous on the open interval (π/2,π/2), the particular solution with arbitrary constant value C is continuous on the interval where π/2<xC<π/2; that is, Cπ/2<x<C+π/2. This situation is fairly typical of nonlinear differential equations, in contrast with linear differential equations, whose solutions are continuous wherever the coefficient functions in the equation are continuous.

FIGURE 1.6.1.

Slope field and solution curves for y=(x+y+3)2.

Example 1 illustrates the fact that any differential equation of the form

dydx=F(ax+by+c) (6)

can be transformed into a separable equation by use of the substitution v=ax+by+c (see Problem 55). The paragraphs that follow deal with other classes of first-order equations for which there are standard substitutions that are known to succeed.

Homogeneous Equations

A homogeneous first-order differential equation is one that can be written in the form

(7)dydx=F(yx).

If we make the substitutions

(8)v=yx,   y=vx,dydx=v+xdvdx,

then Eq. (7) is transformed into the separable equation

xdvdx=F(v)v.

Thus every homogeneous first-order differential equation can be reduced to an integration problem by means of the substitutions in (8).

Remark

A dictionary definition of “homogeneous” is “of a similar kind or nature.” Consider a differential equation of the form

(*)Axmyndydx=Bxpyq+Cxrys,

whose polynomial coefficient functions are “homogeneous” in the sense that each of their terms has the same total degree, m+n=p+q=r+s=K. If we divide each side of (*) by xK, then the result—because xmyn/xm+n=(y/x)n, and so forth—is the equation

A(yx)ndydx=B(yx)q+C(yx)s,

which evidently can be written (by another division) in the form of Eq. (7). More generally, a differential equation of the form P(x,y)y=Q(x,y) with polynomial coefficients P and Q is homogeneous if the terms in these polynomials all have the same total degree K. The differential equation in the following example is of this form with K=2.

Example 2

Solve the differential equation

2xydydx=4x2+3y2.

Solution

This equation is neither separable nor linear, but we recognize it as a homogeneous equation by writing it in the form

dydx=4x2+3y22xy=2(xy)+32(yx).

The substitutions in (8) then take the form

y=vx,dydx=v+xdvdx,v=yx,and1v=xy.

These yield

v+xdvdx=2v+32v,

and hence

xdvdx=2v+v2=v2+42v;2vv2+4dv=1xdx;ln(v2+4)=ln|x|+ln C.

We apply the exponential function to both sides of the last equation to obtain

v2+4=C|x|;y2x2+4=C|x|;y2+4x2=kx3.

Note that the left-hand side of this equation is necessarily nonnegative. It follows that k>0 in the case of solutions that are defined for x>0, while k<0 for solutions where x<0. Indeed, the family of solution curves illustrated in Fig. 1.6.2 exhibits symmetry about both coordinate axes. Actually, there are positive-valued and negative-valued solutions of the forms y(x)=±kx34x2 that are defined for x>4/k if the constant k is positive, and for x<4/k if k is negative.

FIGURE 1.6.2.

Slope field and solution curves for 2xyy=4x2+3y2.

Example 3

Solve the initial value problem

xdydx=y+x2y2,y(x0)=0,

where x0>0.

Solution

We divide both sides by x and find that

dydx=yx+1(yx)2,

so we make the substitutions in (8); we get

v+xdvdx=v+1v2;11v2dv=1xdx;sin1v=ln x+C.

We need not write ln |x| because x>0 near x=x0>0. Now note that v(x0)=y(x0)/x0=0, so C=sin10ln x0=ln x0. Hence

v=yx=sin(ln xlnx0)=sin(lnxx0).

and therefore

y(x)=xsin(lnxx0)

is the desired particular solution. Figure 1.6.3 shows some typical solution curves. Because of the radical in the differential equation, these solution curves are confined to the indicated triangular region x|y|. You can check that the boundary lines y=x and y=x (for x>0) are singular solution curves that consist of points of tangency with the solution curves found earlier.

FIGURE 1.6.3.

Solution curves for xy=y+x2y2.

Bernoulli Equations

A first-order differential equation of the form

(9)dydx+P(x)y=Q(x)yn

is called a Bernoulli equation. If either n=0 or n=1, then Eq. (9) is linear. Otherwise, as we ask you to show in Problem 56, the substitution

(10)v=y1n

transforms Eq. (9) into the linear equation

dvdx+(1n)P(x)v=(1n)Q(x).

Rather than memorizing the form of this transformed equation, it is more efficient to make the substitution in Eq. (10) explicitly, as in the following examples.

Example 4

If we rewrite the homogeneous equation 2xyy=4x2+3y2 of Example 2 in the form

dydx32xy=2xy,

we see that it is also a Bernoulli equation with P(x)=3/(2x), Q(x)=2x, n=1, and 1n=2. Hence we substitute

v=y2,y=v1/2,anddydx=dydvdvdx=12v1/2dvdx.

This gives

12v1/2dvdx32xv1/2=2xv1/2.

Then multiplication by 2v1/2 produces the linear equation

dvdx3xv=4x

with integrating factor ρ=e(3/x)dx=x3. So we obtain

Dx(x3v)=4x2;x3v=4x+C;x3y2=4x+C;y2=4x2+Cx3.

Example 5

The equation

xdydx+6y=3xy4/3

is neither separable nor linear nor homogeneous, but it is a Bernoulli equation with n=43, 1n=13. The substitutions

v=y1/3,y=v3,anddydx=dydvdvdx=3v4dvdx

transform it into

3xv4dvdx+6v3=3xv4.

Division by 3xv4 yields the linear equation

dvdx2xv=1

with integrating factor ρ=e(2/x)dx=x2. This gives

Dx(x2v)=1x2;x2v=1x+C;v=x+Cx2;

and finally,

y(x)=1(x+Cx2)3.

Example 6

The equation

(11)2xe2ydydx=3x4+e2y

is neither separable, nor linear, nor homogeneous, nor is it a Bernoulli equation. But we observe that y appears only in the combinations e2y and Dx(e2y)=2e2yy. This prompts the substitution

v=e2y,   dvdx=2e2ydydx

that transforms Eq. (11) into the linear equation xv(x)=3x4+v(x); that is,

dvdx1xv=3x3.

After multiplying by the integrating factor ρ=1/x, we find that

1xv=3x2dx=x3+C,soe2y=v=x4+Cx,

and hence

y(x)=12ln |x4+Cx|.

Flight Trajectories

Suppose that an airplane departs from the point (a, 0) located due east of its intended destination—an airport located at the origin (0, 0). The plane travels with constant speed v0 relative to the wind, which is blowing due north with constant speed w. As indicated in Fig. 1.6.4, we assume that the plane’s pilot maintains its heading directly toward the origin.

FIGURE 1.6.4.

The airplane headed for the origin.

Figure 1.6.5 helps us derive the plane’s velocity components relative to the ground. They are

dxdt=v0cosθ=v0xx2+y2,dydt=v0sinθ+w=v0yx2+y2+w.

Hence the trajectory y=f(x) of the plane satisfies the differential equation

(12)dydx=dy/dtdx/dt=1v0x(v0ywx2+y2).

FIGURE 1.6.5.

The components of the velocity vector of the airplane.

If we set

(13)k=wv0,

the ratio of the windspeed to the plane’s airspeed, then Eq. (12) takes the homogeneous form

(14)dydx=yxk[1+(yx)2]1/2.

The substitution y=xv, y=v+xv then leads routinely to

(15)dv1+v2=kxdx.

By trigonometric substitution, or by consulting a table for the integral on the left, we find that

(16)ln (v+1+v2)=kln x+C,

and the initial condition v(a)=y(a)/a=0 yields

(17)C=kln a.

As we ask you to show in Problem 68, the result of substituting (17) in Eq. (16) and then solving for v is

(18)v=12[(xa)k(xa)k].

Because y=xv, we finally obtain

(19)y(x)=a2[(xa)1k(xa)1+k]

for the equation of the plane’s trajectory.

Note that only in the case k<1 (that is, w<v0) does the curve in Eq. (19) pass through the origin, so that the plane reaches its destination. If w=v0 (so that k=1), then Eq. (19) takes the form y(x)=12a(1x2/a2), so the plane’s trajectory approaches the point (0, a/2) rather than (0, 0). The situation is even worse if w>v0 (so k>1)-in this case it follows from Eq. (19) that y+ as x0. The three cases are illustrated in Fig. 1.6.6.

FIGURE 1.6.6.

The three cases w<v0 (plane velocity exceeds wind velocity), w=v0 (equal velocities), and w>v0 (wind is greater).

Example 7

Flight trajectory If a=200 mi, v0=500 mi/h, and w=100 mi/h, then k=w/v0=15, so the plane will succeed in reaching the airport at (0, 0). With these values, Eq. (19) yields

(20)y(x)=100[(x200)4/5(x200)6/5].

Now suppose that we want to find the maximum amount by which the plane is blown off course during its trip. That is, what is the maximum value of y(x) for 0x200?

Solution

Differentiation of the function in Eq. (20) yields

dydx=12[45(x200)1/565(x200)1/5],

and we readily solve the equation y(x)=0 to obtain (x/200)2/5=23. Hence

ymax=100[(23)2(23)3]=4002714.81.

Thus the plane is blown almost 15 mi north at one point during its westward journey. (The graph of the function in Eq. (20) is the one used to construct Fig. 1.6.4. The vertical scale there is exaggerated by a factor of 4.)

Exact Differential Equations

We have seen that a general solution y(x) of a first-order differential equation is often defined implicitly by an equation of the form

(21)F(x,y(x))=C,

where C is a constant. On the other hand, given the identity in (21), we can recover the original differential equation by differentiating each side with respect to x. Provided that Eq. (21) implicitly defines y as a differentiable function of x, this gives the original differential equation in the form

Fx+Fydydx=0;

that is,

(22)M(x,y)+N(x,y)dydx=0,

where M(x,y)=Fx(x,y) and N(x,y)=Fy(x,y).

It is sometimes convenient to rewrite Eq. (22) in the more symmetric form

(23)M(x,y)dx+N(x,y)dy=0,

called its differential form. The general first-order differential equation y=f(x,y) can be written in this form with M=f(x,y) and N1. The preceding discussion shows that, if there exists a function F(x,y) such that

Fx=MandFy=N,

then the equation

F(x,y)=C

implicitly defines a general solution of Eq. (23). In this case, Eq. (23) is called an exact differential equation—the differential

dF=Fxdx+Fydy

of F(x,y) is exactly M dx+N dy.

Natural questions are these: How can we determine whether the differential equation in (23) is exact? And if it is exact, how can we find the function F such that Fx=M and Fy=N? To answer the first question, let us recall that if the mixed second-order partial derivatives Fxy and Fyx are continuous on an open set in the xy-plane, then they are equal: Fxy=Fyx. If Eq. (23) is exact and M and N have continuous partial derivatives, it then follows that

My=Fxy=Fyx=Nx.

Thus the equation

(24)My=Nx

is a necessary condition that the differential equation M dx+N dy=0 be exact. That is, if MyNx, then the differential equation in question is not exact, so we need not attempt to find a function F(x, y) such that Fx=M and Fy=N—there is no such function.

Example 8

The differential equation

(25)y3dx+3xy2dy=0

is exact because we can immediately see that the function F(x,y)=xy3 has the property that Fx=y3 and Fy=3xy2. Thus a general solution of Eq. (25) is

xy3=C;

if you prefer, y(x)=kx1/3.

But suppose that we divide each term of the differential equation in Example 8 by y2 to obtain

(26)y dx+3x dy=0.

This equation is not exact because, with M=y and N=3x, we have

My=13=Nx.

Hence the necessary condition in Eq. (24) is not satisfied.

We are confronted with a curious situation here. The differential equations in (25) and (26) are essentially equivalent, and they have exactly the same solutions, yet one is exact and the other is not. In brief, whether a given differential equation is exact or not is related to the precise form M dx+N dy=0 in which it is written.

Theorem 1 tells us that (subject to differentiability conditions usually satisfied in practice) the necessary condition in (24) is also a sufficient condition for exactness. In other words, if My=Nx, then the differential equation M dx+N dy=0 is exact.

Proof:

We have seen already that it is necessary for Eq. (24) to hold if Eq. (23) is to be exact. To prove the converse, we must show that if Eq. (24) holds, then we can construct a function F(x, y) such that F/x=M and F/y=N. Note first that, for any function g(y), the function

(27)F(x, y)=M(x, y)dx+g(y)

satisfies the condition F/x=M. (In Eq. (27), the notation M(x, y)dx denotes an antiderivative of M(x, y) with respect to x.) We plan to choose g(y) so that

N=Fy=(yM(x, y)dx)+g(y)

as well; that is, so that

(28)g(y)=NyM(x, y)dx.

To see that there is such a function of y, it suffices to show that the right-hand side in Eq. (28) is a function of y alone. We can then find g(y) by integrating with respect to y. Because the right-hand side in Eq. (28) is defined on a rectangle, and hence on an interval as a function of x, it suffices to show that its derivative with respect to x is identically zero. But

x(NyM(x, y)dx)=NxxyM(x, y)dx=NxyxM(x, y)dx=NxMy=0

by hypothesis. So we can, indeed, find the desired function g(y) by integrating Eq. (28). We substitute this result in Eq. (27) to obtain

(29)F(x, y)=M(x, y)dx+(N(x, y)yM(x, y)dx)dy

as the desired function with Fx=M and Fy=N.

Instead of memorizing Eq. (29), it is usually better to solve an exact equation M dx+N dy=0 by carrying out the process indicated by Eqs. (27) and (28). First we integrate M(x, y) with respect to x and write

F(x, y)=M(x, y)dx+g(y),

thinking of the function g(y) as an “arbitrary constant of integration” as far as the variable x is concerned. Then we determine g(y) by imposing the condition that F/y=N(x, y). This yields a general solution in the implicit form F(x, y)=C.

Example 9

Solve the differential equation

(30)(6xyy3)dx+(4y+3x23xy2)dy=0.

Solution

Let M(x, y)=6xyy3 and N(x, y)=4y+3x23xy2. The given equation is exact because

My=6x3y2=Nx.

Integrating F/x=M(x, y) with respect to x, we get

F(x, y)=(6xyy3)dx=3x2yxy3+g(y).

Then we differentiate with respect to y and set F/y=N(x, y). This yields

Fy=3x23xy2+g(y)=4y+3x23xy2,

and it follows that g(y)=4y. Hence g(y)=2y2+C1, and thus

F(x, y)=3x2yxy3+2y2+C1.

Therefore, a general solution of the differential equation is defined implicitly by the equation

(31)3x2yxy3+2y2=C

(we have absorbed the constant C1 into the constant C).

Remark

Figure 1.6.7 shows a rather complicated structure of solution curves for the differential equation of Example 9. The solution satisfying a given initial condition y(x0)=y0 is defined implicitly by Eq. (31), with C determined by substituting x=x0 and y=y0 in the equation. For instance, the particular solution satisfying y(0)=1 is defined implicitly by the equation 3x2yxy3+2y2=2. The other two special points in the figure—at (0, 0) and near (0.75, 2.12)—are ones where both coefficient functions in Eq. (30) vanish, so the theorem of Section 1.3 does not guarantee a unique solution.

FIGURE 1.6.7.

Slope field and solution curves for the exact equation in Example 9.

Reducible Second-Order Equations

A second-order differential equation involves the second derivative of the unknown function y(x), and thus has the general form

(32)F(x, y, y, y)=0.

If either the dependent variable y or the independent variable x is missing from a second-order equation, then it is easily reduced by a simple substitution to a first-order equation that may be solvable by the methods of this chapter.

Dependent variable y missing. If y is missing, then Eq. (32) takes the form

(33)F(x, y, y)=0.

Then the substitution

(34)p=y=dydx,y=dpdx

results in the first-order differential equation

F(x, p, p)=0.

If we can solve this equation for a general solution p(x, C1) involving an arbitrary constant C1, then we need only write

x(x)=y(x)dx=p(x, C1)dx+C2

to get a solution of Eq. (33) that involves two arbitrary constants C1 and C2 (as is to be expected in the case of a second-order differential equation).

Example 10

Solve the equation xy+2y=6x in which the dependent variable y is missing.

Solution

The substitution defined in (34) gives the first-order equation

xdpdx+2p=6x;that is,dpdx+2xp=6.

Observing that the equation on the right here is linear, we multiply by its integrating factor ρ=exp((2/x)dx)=e2 ln x=x2 and get

Dx(x2p)=6x2,x2p=2x3+C1,p=dydx=2x+C1x2.

A final integration with respect to x yields the general solution

y(x)=x2C1x+C2

of the second-order equation xy+2y=6x. Solution curves with C1=0 but C20 are simply vertical translates of the parabola y=x2 (for which C1=C2=0). Figure 1.6.8 shows this parabola and some typical solution curves with C2=0 but C10. Solution curves with C1 and C2 both nonzero are vertical translates of those (other than the parabola) shown in Fig. 1.6.8.

FIGURE 1.6.8.

Solution curves of the form y(x)=x2C1x for C1=0, ±3, ±10, ±20, ±35, ±60, ±100.

Independent variable x missing. If x is missing, then Eq. (32) takes the form

(35)F(y, y, y)=0.

Then the substitution

(36)p=y=dydx,y=dpdx=dpdydydx=pdpdy

results in the first-order differential equation

F(y, p, pdpdy)=0

for p as a function of y. If we can solve this equation for a general solution p(y,C1) involving an arbitrary constant C1, then (assuming that y0) we need only write

x(y)=dxdydy=1dy/dxdy=1pdy=dyp(y, C1)+C2.

If the final integral P=(1/p)dy can be evaluated, the result is an implicit solution x(y)=P(y, C1)+C2 of our second-order differential equation.

Example 11

Solve the equation yy=(y)2 in which the independent variable x is missing.

Solution

We assume temporarily that y and y are both nonnegative, and then point out at the end that this restriction is unnecessary. The substitution defined in (36) gives the first-order equation

ypdpdy=p2.

Then separation of variables gives

dpp=dyy,ln p=ln y+C(because y>0 and p=y>0),p=C1y

where C1=eC. Hence

dxdy=1p=1C1y,C1x=dyy=ln y+C2.

The resulting general solution of the second-order equation yy=(y)2 is

y(x)=exp(C1xC2)=AeBx,

where A=eC2 and B=C1. Despite our temporary assumptions, which imply that the constants A and B are both positive, we readily verify that y(x)=AeBx satisfies yy=(y)2 for all real values of A and B. With B=0 and different values of A, we get all horizontal lines in the plane as solution curves. The upper half of Fig. 1.6.9 shows the solution curves obtained with A=1 (for instance) and different positive values of B. With A=1 these solution curves are reflected in the x-axis, and with negative values of B they are reflected in the y-axis. In particular, we see that we get solutions of yy=(y)2, allowing both positive and negative possibilities for both y and y.

FIGURE 1.6.9.

The solution curves y=AeBx with B=0 and A=0, ±1 are the horizontal lines y=0, ±1. The exponential curves with B>0 and A=±1 are in color, those with B<0 and A=±1 are black.

1.6 Problems

Find general solutions of the differential equations in Problems 1 through 30. Primes denote derivatives with respect to x throughout.

  1. (x+y)y=xy

  2. 2xyy=x2+2y2

  3. xy=y+2xy

  4. (xy)y=x+y

  5. x(x+y)y=y(xy)

  6. (x+2y)y=y

  7. xy2y=x3+y3

  8. x2y=xy+x2ey/x

  9. x2y=xy+y2

  10. xyy=x2+3y2

  11. (x2y2)y=2xy

  12. xyy=y2+x4x2+y2

  13. xy=y+x2+y2

  14. yy+x=x2+y2

  15. x(x+y)y+y(3x+y)=0

  16. y=x+y+1

  17. y=(4x+y)2

  18. (x+y)y=1

  19. x2y+2xy=5y3

  20. y2y+2xy3=6x

  21. y=y+y3

  22. x2y+2xy=5y4

  23. xy+6y=3xy4/3

  24. 2xy+y3e2x=2xy

  25. y2(xy+y)(1+x4)1/2=x

  26. 3y2y+y3=ex

  27. 3xy2y=3x4+y3

  28. xeyy=2(ey+x3e2x)

  29. (2xsin ycos y)y=4x2+sin2 y

  30. (x+ey)y=xey1

In Problems 31 through 42, verify that the given differential equation is exact; then solve it.

  1. (2x+3y)dx+(3x+2y)dy=0

  2. (4xy)dx+(6yx)dy=0

  3. (3x2+2y2)dx+(4xy+6y2)dy=0

  4. (2xy2+3x2)dx+(2x2y+4y3)dy=0

  5. (x3+yx)dx+(y2+ln x)dy=0

  6. (1+yexy)dx+(2y+xexy)dy=0

  7. (cos x+ln y)dx+(xy+ey)dy=0

  8. (x+tan1 y)dx+x+y1+y2dy=0

  9. (3x2y3+y4)dx+(3x3y2+y4+4xy3)dy=0

  10. (ex sin y+tan y)dx+(ex cos y+xsec2 y)dy=0

  11. (2xy3y2x4)dx+(2yx3x2y2+1y)dy=0

  12. 2x5/23y5/32x5/2y2/3dx+3y5/32x5/23x3/2y5/3dy=0

Find a general solution of each reducible second-order differential equation in Problems 43–54. Assume x, y and/or y positive where helpful (as in Example 11).

  1. xy=y

  2. yy+(y)2=0

  3. y+4y=0

  4. xy+y=4x

  5. y=(y)2

  6. x2y+3xy=2

  7. yy+(y)2=yy

  8. y=(x+y)2

  9. y=2y(y)3

  10. y3y=1

  11. y=2yy

  12. yy=3(y)2

  13. Show that the substitution v=ax+by+c transforms the differential equation dy/dx=F(ax+by+c) into a separable equation.

  14. Suppose that n0 and n1. Show that the substitution v=y1n transforms the Bernoulli equation dy/dx+P(x)y=Q(x)yn into the linear equation

    dvdx+(1n)P(x)v(x)=(1n)Q(x).
  15. Show that the substitution v=ln y transforms the differential equation dy/dx+P(x)y=Q(x)(y ln y) into the linear equation dv/dx+P(x)=Q(x)v(x).

  16. Use the idea in Problem 57 to solve the equation

    xdydx4x2y+2yln y=0.
  17. Solve the differential equation

    dydx=xy1x+y+3

    by finding h and k so that the substitutions x=u+h, y=v+k transform it into the homogeneous equation

    dvdu=uvu+v.
  18. Use the method in Problem 59 to solve the differential equation

    dydx=2xx+74x3y18.
  19. Make an appropriate substitution to find a solution of the equation dy/dx=sin(xy). Does this general solution contain the linear solution y(x)=xπ/2 that is readily verified by substitution in the differential equation?

  20. Show that the solution curves of the differential equation

    dydx=y(2x3y3)x(2y3x3)

    are of the form x3+y3=Cxy.

  21. The equation dy/dx=A(x)y2+B(x)y+C(x) is called a Riccati equation. Suppose that one particular solution y1(x) of this equation is known. Show that the substitution

    y=y1+1v

    transforms the Riccati equation into the linear equation

    dvdx+(B+2Ay1)v=A.

Use the method of Problem 63 to solve the equations in Problems 64 and 65, given that y1(x)=x is a solution of each.

  1. dydx+y2=1+x2

  2. dydx+2xy=1+x2+y2

  3. An equation of the form

    (37)y=xy+g(y)

    is called a Clairaut equation. Show that the one-parameter family of straight lines described by

    (38)y(x)=Cx+g(C)

    is a general solution of Eq. (37).

  4. Consider the Clairaut equation

    y=xy14(y)2

    for which g(y)=14(y)2 in Eq. (37). Show that the line

    y=Cx14C2

    is tangent to the parabola y=x2 at the point (12C, 14C2).

    Explain why this implies that y=x2 g is a singular solution of the given Clairaut equation. This singular solution and the one-parameter family of straight line solutions are illustrated in Fig. 1.6.10.

    FIGURE 1.6.10.

    Solutions of the Clairaut equation of Problem 67. The “typical” straight line with equation y=Cx14C2 is tangent to the parabola at the point (12C, 14C2).

  5. Derive Eq. (18) in this section from Eqs. (16) and (17).

  6. Flight trajectory In the situation of Example 7, suppose that a=100 mi,v0=400 mi/h, and w=40 mi/h. Now how far northward does the wind blow the airplane?

  7. Flight trajectory As in the text discussion, suppose that an airplane maintains a heading toward an airport at the origin. If v0=500 mi/h and w=50 mi/h (with the wind blowing due north), and the plane begins at the point (200, 150), show that its trajectory is described by

    y+x2+y2=2(200x9)1/10.
  8. River crossing A river 100 ft wide is flowing north at w feet per second. A dog starts at (100, 0) and swims at v0=4 ft/s, always heading toward a tree at (0, 0) on the west bank directly across from the dog’s starting point. (a) If w=2 ft/s, show that the dog reaches the tree. (b) If w=4 ft/s, show that the dog reaches instead the point on the west bank 50 ft north of the tree. (c) If w=6 ft/s, show that the dog never reaches the west bank.

  9. In the calculus of plane curves, one learns that the curvature κ of the curve y=y(x) at the point (x, y) is given by

    κ=|y(x)|[1+y(x)2]3/2,

    and that the curvature of a circle of radius r is κ=1/r. [See Example 3 in Section 11.6 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008.] Conversely, substitute ρ=y to derive a general solution of the second-order differential equation

    ry=[1+(y)2]3/2

    (with r constant) in the form

    (xa)2+(yb)2=r2.

    Thus a circle of radius r (or a part thereof) is the only plane curve with constant curvature 1/r.

1.6 Application Computer Algebra Solutions

Computer algebra systems typically include commands for the “automatic” solution of differential equations. But two different such systems often give different results whose equivalence is not clear, and a single system may give the solution in an overly complicated form. Consequently, computer algebra solutions of differential equations often require considerable “processing” or simplification by a human user in order to yield concrete and applicable information. Here we illustrate these issues using the interesting differential equation

(1)dydx=sin(xy)

that appeared in the Section 1.3 Application. The Maple command

dsolve( D(y)(x) = sin(x - y(x)), y(x));

yields the simple and attractive result

(2)y(x)=x2 tan1(x2C1xC1)

that was cited there. But the supposedly equivalent Mathematica command

DSolve[ y′[x] == Sin[x y[x]], y[x], x]

and the Wolfram|Alpha query

Y′ = sin(x - y)

both yield considerably more complicated results from which—with a fair amount of effort in simplification—one can extract the quite different looking solution

(3)y(x)=2 cos1(2 cosx2+(xc)(cosx2+sinx2)2+2(xc+1)2).

This apparent disparity is not unusual; different symbolic algebra systems, or even different versions of the same system, often yield different forms of a solution of the same differential equation. As an alternative to attempted reconciliation of such seemingly disparate results as in Eqs. (2) and (3), a common tactic is simplification of the differential equation before submitting it to a computer algebra system.

Exercise 1: Show that the plausible substitution v=xy in Eq. (1) yields the separable equation

(4)dvdx=1sin v.

Now the Maple command int(1/(1 - sin(v)), v) yields

(5)dv1sin v=21tanv2

(omitting the constant of integration, as symbolic computer algebra systems often do).

Exercise 2: Use simple algebra to deduce from Eq. (5) the integral formula

(6)dv1sin v=1+tanv21tanv2+C.

Exercise 3: Deduce from (6) that Eq. (4) has the general solution

v(x)=2 tan1(x1+Cx+1+C),

and hence that Eq. (1) has the general solution

(7)y(x)=x2 tan1(x1+Cx+1+C).

Exercise 4: Finally, reconcile the forms in Eq. (2) and Eq. (7). What is the relation between the constants C and C1?

Exercise 5: Show that the integral in Eq. (5) yields immediately the graphing calculator implicit solution shown in Fig. 1.6.11.

Investigation: For your own personal differential equation, let p and q be two distinct nonzero digits in your student ID number, and consider the differential equation

FIGURE 1.6.11.

Implicit solution of y=sin(xy) generated by a TI-Nspire CX CAS.

(8)dydx=1pcos(xqy).
  1. Find a symbolic general solution using a computer algebra system and/or some combination of the techniques listed in this project.

  2. Determine the symbolic particular solution corresponding to several typical initial conditions of the form y(x0)=y0.

  3. Determine the possible values of a and b such that the straight line y=ax+b is a solution curve of Eq. (8).

  4. Plot a direction field and some typical solution curves. Can you make a connection between the symbolic solution and your (linear and nonlinear) solution curves?

Chapter 1 Summary

In this chapter we have discussed applications of and solution methods for several important types of first-order differential equations, including those that are separable (Section 1.4), linear (Section 1.5), or exact (Section 1.6). In Section 1.6 we also discussed substitution techniques that can sometimes be used to transform a given first-order differential equation into one that is either separable, linear, or exact.

Lest it appear that these methods constitute a “grab bag” of special and unrelated techniques, it is important to note that they are all versions of a single idea. Given a differential equation

(1)f(x, y, y)=0,

we attempt to write it in the form

(2)ddx[G(x, y)]=0.

It is precisely to obtain the form in Eq. (2) that we multiply the terms in Eq. (1) by an appropriate integrating factor (even if all we are doing is separating the variables). But once we have found a function G(x, y) such that Eqs. (1) and (2) are equivalent, a general solution is defined implicitly by means of the equation

(3)G(x, y)=C

that one obtains by integrating Eq. (2).

Given a specific first-order differential equation to be solved, we can attack it by means of the following steps:

Many first-order differential equations succumb to the line of attack outlined here. Nevertheless, many more do not. Because of the wide availability of computers, numerical techniques are commonly used to approximate the solutions of differential equations that cannot be solved readily or explicitly by the methods of this chapter. Indeed, most of the solution curves shown in figures in this chapter were plotted using numerical approximations rather than exact solutions. Several numerical methods for the appropriate solution of differential equations will be discussed in Chapter 2.

Chapter 1 Review Problems

Find general solutions of the differential equations in Problems 1 through 30. Primes denote derivatives with respect to x.

  1. x3+3yxy=0

  2. xy2+3y2x2y=0

  3. xy+y2x2y=0

  4. 2xy3+ex+(3x2y2+sin y)y=0

  5. 3y+x4y=2xy

  6. 2xy2+x2y=y2

  7. 2x2y+x3y=1

  8. 2xy+x2y=y2

  9. xy+2y=6x2y

  10. y=1+x2+y2+x2y2

  11. x2y=xy+3y2

  12. 6xy3+2y4+(9x2y2+8xy3)y=0

  13. 4xy2+y=5x4y2

  14. x3y=x2yy3

  15. y+3y=3x2e3x

  16. y=x22xy+y2

  17. ex+yexy+(ey+xeyx)y=0

  18. 2x2yx3y=y3

  19. 3x5y2+x3y=2y2

  20. xy+3y=3x3/2

  21. (x21)y+(x1)y=1

  22. xy=6y+12x4y2/3

  23. ey+ycos x+(xey+sin x)y=0

  24. 9x2y2+x3/2y=y2

  25. 2y+(x+1)y=3x+3

  26. 9x1/2y4/312x1/5y3/2+(8x3/2y1/315x6/5y1/2)y=0

  27. 3y+x3y4+3xy=0

  28. y+xy=2e2x

  29. (2x+1)y+y=(2x+1)3/2

  30. y=x+y

Each of the differential equations in Problems 31 through 36 is of two different types considered in this chapter—separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results.

  1. dydx=3(y+7)x2

  2. dydx=xy3xy

  3. dydx=3x2+2y24xy

  4. dydx=x+3yy3x

  5. dydx=2xy+2xx2+1

  6. dydx=yytan x

2 Mathematical Models and Numerical Methods

2.1 Population Models

In Section 1.4 we introduced the exponential differential equation dP/dt=kP, with solution P(t)=P0ekt, as a mathematical model for natural population growth that occurs as a result of constant birth and death rates. Here we present a more general population model that accommodates birth and death rates that are not necessarily constant. As before, however, our population function P(t) will be a continuous approximation to the actual population, which of course changes only by integral increments—that is, by one birth or death at a time.

Suppose that the population changes only by the occurrence of births and deaths—there is no immigration or emigration from outside the country or environment under consideration. It is customary to track the growth or decline of a population in terms of its birth rate and death rate functions defined as follows:

  • β(t) is the number of births per unit of population per unit of time at time t;

  • δ(t) is the number of deaths per unit of population per unit of time at time t.

Then the numbers of births and deaths that occur during the time interval [t,t+Δt] is given (approximately) by

births:β(t)P(t)Δt,deaths:δ(t)P(t)Δt.

Hence the change ΔP in the population during the time interval [t,t+Δt] of length Δt is

ΔP={births}{deaths}β(t)P(t)Δtδ(t)P(t)Δt,

so

ΔPΔt[β(t)δ(t)]P(t).

The error in this approximation should approach zero as Δt0, so—taking the limit—we get the differential equation

(1)dPdt=(βδ)P,

in which we write β=β(t), δ=δ(t), and P=P(t) for brevity. Equation (1) is the general population equation. If β and δ are constant, Eq. (1) reduces to the natural growth equation with k=βδ. But it also includes the possibility that β and δ are variable functions of t. The birth and death rates need not be known in advance; they may well depend on the unknown function P(t).

Example 1

Population explosion Suppose that an alligator population numbers 100 initially, and that its death rate is δ=0 (so none of the alligators is dying). If the birth rate is β=(0.0005)P—and thus increases as the population does—then Eq. (1) gives the initial value problem

dPdt=(0.0005)P2,P(0)=100

(with t in years). Then upon separating the variables we get

1P2dP=(0.0005)dt;1P=(0.0005)t+C.

Substitution of t=0, P=100 gives C=1/100, and then we readily solve for

P(t)=200020t.

For instance, P(10)=2000/10=200, so after 10 years the alligator population has doubled. But we see that P+ as t20, so a real “population explosion” occurs in 20 years. Indeed, the direction field and solution curves shown in Fig. 2.1.1 indicate that a population explosion always occurs, whatever the size of the (positive) initial population P(0)=P0. In particular, it appears that the population always becomes unbounded in a finite period of time.

FIGURE 2.1.1.

Slope field and solution curves for the equation dP/dt=(0.0005)P2 in Example 1.

Bounded Populations and the Logistic Equation

In situations as diverse as the human population of a nation and a fruit fly population in a closed container, it is often observed that the birth rate decreases as the population itself increases. The reasons may range from increased scientific or cultural sophistication to a limited food supply. Suppose, for example, that the birth rate β is a linear decreasing function of the population size P, so that β=β0β1P, where β0 and β1 are positive constants. If the death rate δ=δ0 remains constant, then Eq. (1) takes the form

dPdt=(β0β1Pδ0)P;

that is,

(2)dPdt=aPbP2,

where a=β0δ0 and b=β1.

If the coefficients a and b are both positive, then Eq. (2) is called the logistic equation. For the purpose of relating the behavior of the population P(t) to the values of the parameters in the equation, it is useful to rewrite the logistic equation in the form

(3)dPdt=kP(MP),

where k=b and M=a/b are constants.

Example 2

Logistic model In Example 4 of Section 1.3 we explored graphically a population that is modeled by the logistic equation

(4)dPdt=0.0004P(150P)=0.06P0.0004P2.

To solve this differential equation symbolically, we separate the variables and integrate. We get

dPP(150P)=0.0004 dt,1150(1P+1150P) dP=0.0004 dt[partial fractions],ln |P|ln |150  P|=0.06t+C,P150P=±eCe0.06t=Be0.06t[where B=±eC].

If we substitute t=0 and P=P0150 into this last equation, we find that B=P0/(150P0). Hence

P150P=P0e0.06t150P0.

Finally, this equation is easy to solve for the population

(5)P(t)=150P0P0+(150P0)e0.06t

at time t in terms of the initial population P0=P(0). Figure 2.1.2 shows a number of solution curves corresponding to different values of the initial population ranging from P0=20 to P0=300. Note that all these solution curves appear to approach the horizontal line P=150 as an asymptote. Indeed, you should be able to see directly from Eq. (5) that limtP(t)=150, whatever the initial value P0>0.

FIGURE 2.1.2.

Typical solution curves for the logistic equation P=0.06P0.0004P2.

Limiting Populations and Carrying Capacity

The finite limiting population noted in Example 2 is characteristic of logistic populations. In Problem 32 we ask you to use the method of solution of Example 2 to show that the solution of the logistic initial value problem

(6)dPdt=kP(MP),P(0)=P0

is

(7)P(t)=MP0P0+(MP0)ekMt.

Actual animal populations are positive valued. If P0=M, then (7) reduces to the unchanging (constant-valued) “equilibrium population” P(t)M. Otherwise, the behavior of a logistic population depends on whether 0<P0<M or P0>M. If 0<P0<M, then we see from (6) and (7) that P>0 and

P(t)=MP0P0+(MP0)ekMt=MP0P0+{pos. number}<MP0P0=M.

However, if P0>M, then we see from (6) and (7) that P<0 and

P(t)=MP0P0+(MP0)ekMt=MP0P0+{neg. number}>MP0P0=M.

In either case, the “positive number” or “negative number” in the denominator has absolute value less than P0 and—because of the exponential factor—approaches 0 as t+. It follows that

(8)limt+P(t)=MP0P0+0=M.

Thus a population that satisfies the logistic equation does not grow without bound like a naturally growing population modeled by the exponential equation P=kP. Instead, it approaches the finite limiting population M as t+. As illustrated by the typical logistic solution curves in Fig. 2.1.3, the population P(t) steadily increases and approaches M from below if 0<P0<M, but steadily decreases and approaches M from above if P0>M. Sometimes M is called the carrying capacity of the environment, considering it to be the maximum population that the environment can support on a long-term basis.

FIGURE 2.1.3.

Typical solution curves for the logistic equation P=kP(MP). Each solution curve that starts below the line P=M/2 has an inflection point on this line. (See Problem 34.)

Example 3

Limiting population Suppose that in 1885 the population of a certain country was 50 million and was growing at the rate of 750,000 people per year at that time. Suppose also that in 1940 its population was 100 million and was then growing at the rate of 1 million per year. Assume that this population satisfies the logistic equation. Determine both the limiting population M and the predicted population for the year 2000.

Solution

We substitute the two given pairs of data in Eq. (3) and find that

0.75=50k(M50),1.00=100k(M100).

We solve simultaneously for M=200 and k=0.0001. Thus the limiting population of the country in question is 200 million. With these values of M and k, and with t=0 corresponding to the year 1940 (in which P0=100), we find that—according to Eq. (7)—the population in the year 2000 will be

P(60)=100200100+(200100)e(0.0001)(200)(60),

about 153.7 million people.

Historical Note

The logistic equation was introduced (around 1840) by the Belgian mathematician and demographer P. F. Verhulst as a possible model for human population growth. In the next two examples we compare natural growth and logistic model fits to the 19th-century U.S. population census data, then compare projections for the 20th century.

Example 4

Percentage growth rate The U.S. population in 1800 was 5.308 million and in 1900 was 76.212 million. If we take P0=5.308 (with t=0 in 1800) in the natural growth model P(t)=P0ert and substitute t=100, P=76.212, we find that

76.212=5.308e100r,sor=1100ln76.2125.3080.026643.

Thus our natural growth model for the U.S. population during the 19th century is

(9)P(t)=(5.308)e(0.026643)t

(with t in years and P in millions). Because e0.0266431.02700, the average population growth between 1800 and 1900 was about 2.7% per year.

Example 5

Logistic modeling The U.S. population in 1850 was 23.192 million. If we take P0=5.308 and substitute the data pairs t=50, P=23.192 (for 1850) and t=100, P=76.212 (for 1900) in the logistic model formula in Eq. (7), we get the two equations

(10)(5.308)M5.308+(M5.308)e50kM=23.192,(5.308)M5.308+(M5.308)e100kM=76.212

in the two unknowns k and M. Nonlinear systems like this ordinarily are solved numerically using an appropriate computer system. But with the right algebraic trick (Problem 36 in this section) the equations in (10) can be solved manually for k=0.000167716, M=188.121. Substitution of these values in Eq. (7) yields the logistic model

(11)P(t)=998.5465.308+(182.813)e(0.031551)t.

The table in Fig. 2.1.4 compares the actual 1800–2010 U.S. census population figures with those predicted by the exponential growth model in (9) and by the logistic model in (11). Both models agree well with the 19th-century figures. But the exponential model diverges appreciably from the census data in the early decades of the 20th century, whereas the logistic model remains accurate until 1940. By the end of the 20th century the exponential model vastly overestimates the actual U.S. population; indeed it predicts a U.S. population of nearly 1.5 billion in the year 2010, over 3.6 times the actual value. The logistic model, on the other hand, underestimates the U.S. population, but with a percentage error of less than 43%.

FIGURE 2.1.4.

Comparison of exponential growth and logistic models with U.S. census populations (in millions).

Year Actual U.S. Pop. Exponential Model Exponential Error Logistic Model Logistic Error
1800 5.308 5.308 0.000 5.308 0.000
1810 7.240 6.929 0.311 7.202 0.038
1820 9.638 9.044 0.594 9.735 0.097
1830 12.861 11.805 1.056 13.095 0.234
1840 17.064 15.409 1.655 17.501 0.437
1850 23.192 20.113 3.079 23.192 0.000
1860 31.443 26.253 5.190 30.405 1.038
1870 38.558 34.268 4.290 39.326 0.768
1880 50.189 44.730 5.459 50.034 0.155
1890 62.980 58.387 4.593 62.435 0.545
1900 76.212 76.212 0.000 76.213 0.001
1910 92.228 99.479 7.251 90.834 1.394
1920 106.022 129.849 23.827 105.612 0.410
1930 123.203 169.492 46.289 119.834 3.369
1940 132.165 221.237 89.072 132.886 0.721
1950 151.326 288.780 137.454 144.354 6.972
1960 179.323 376.943 197.620 154.052 25.271
1970 203.302 492.023 288.721 161.990 41.312
1980 226.542 642.236 415.694 168.316 58.226
1990 248.710 838.308 589.598 173.252 75.458
2000 281.422 1094.240 812.818 177.038 104.384
2010 308.745 1428.307 1119.562 179.905 128.839

The two models are compared in Fig. 2.1.5, where plots of their respective errors—as a percentage of the actual population—are shown for the 1800–1950 period. We see that the logistic model tracks the actual population reasonably well throughout this 150-year period. However, the exponential error is considerably larger during the 19th century and literally goes off the chart during the first half of the 20th century.

FIGURE 2.1.5.

Percentage errors in the exponential and logistic population models for 1800–1950.

In order to measure the extent to which a given model fits actual data, it is customary to define the average error (in the model) as the square root of the average of the squares of the individual errors (the latter appearing in the fourth and sixth columns of the table in Fig. 2.1.4). Using only the 1800–1900 data, this definition gives 3.162 for the average error in the exponential model, while the average error in the logistic model is only 0.452. Consequently, even in 1900 we might well have anticipated that the logistic model would predict the U.S. population growth during the 20th century more accurately than the exponential model.

The moral of Examples 4 and 5 is simply that one should not expect too much of models that are based on severely limited information (such as just a pair of data points). Much of the science of statistics is devoted to the analysis of large “data sets” to formulate useful (and perhaps reliable) mathematical models.

More Applications of the Logistic Equation

We next describe some situations that illustrate the varied circumstances in which the logistic equation is a satisfactory mathematical model.

  1. Limited environment situation. A certain environment can support a population of at most M individuals. It is then reasonable to expect the growth rate βδ (the combined birth and death rates) to be proportional to MP, because we may think of MP as the potential for further expansion. Then βδ=k(MP), so that

    dPdt=(βδ)P=kP(MP).

    The classic example of a limited environment situation is a fruit fly population in a closed container.

  2. Competition situation. If the birth rate β is constant but the death rate δ is proportional to P, so that δ=αP, then

    dPdt=(βαP)P=kP(MP).

    This might be a reasonable working hypothesis in a study of a cannibalistic population, in which all deaths result from chance encounters between individuals. Of course, competition between individuals is not usually so deadly, nor its effects so immediate and decisive.

  3. Joint proportion situation. Let P(t) denote the number of individuals in a constant-size susceptible population M who are infected with a certain contagious and incurable disease. The disease is spread by chance encounters. Then P(t) should be proportional to the product of the number P of individuals having the disease and the number MP of those not having it, and therefore dP/dt=kP(MP). Again we discover that the mathematical model is the logistic equation. The mathematical description of the spread of a rumor in a population of M individuals is identical.

Example 6

Spread of rumor Suppose that at time t=0, 10 thousand people in a city with population M=100 thousand people have heard a certain rumor. After 1 week the number P(t) of those who have heard it has increased to P(1)=20 thousand. Assuming that P(t) satisfies a logistic equation, when will 80% of the city’s population have heard the rumor?

Solution

Substituting P0=10 and M=100 (thousand) in Eq. (7), we get

(12)P(t)=100010+90e100kt.

Then substitution of t=1, P=20 gives the equation

20=100010+90e100k

that is readily solved for

e100k=49,sok=1100ln940.008109.

With P(t)=80, Eq. (12) takes the form

80=100010+90e100kt,

which we solve for e100kt=136. It follows that 80% of the population has heard the rumor when

t=ln 36100k=ln 36ln944.42,

thus after about 4 weeks and 3 days.

Doomsday versus Extinction

Consider a population P(t) of unsophisticated animals in which females rely solely on chance encounters to meet males for reproductive purposes. It is reasonable to expect such encounters to occur at a rate that is proportional to the product of the number P/2 of males and the number P/2 of females, hence at a rate proportional to P2. We therefore assume that births occur at the rate kP2 (per unit time, with k constant). The birth rate (births/time/population) is then given by β=kP. If the death rate δ is constant, then the general population equation in (1) yields the differential equation

(13)dPdt=kP2δP=kP(PM)

(where M=δ/k>0) as a mathematical model of the population.

Note that the right-hand side in Eq. (13) is the negative of the right-hand side in the logistic equation in (3). We will see that the constant M is now a threshold population, with the way the population behaves in the future depending critically on whether the initial population P0 is less than or greater than M.

Example 7

Doomsday vs. extinction Consider an animal population P(t) that is modeled by the equation

(14)dPdt=0.0004P(P150)=0.0004P20.06P.

We want to find P(t) if (a) P(0)=200; (b) P(0)=100.

Solution

To solve the equation in (14), we separate the variables and integrate. We get

(15)dPP(P150)=0.0004 dt,1150(1P1P150)dP=0.0004 dt[partial fractions],ln |P|ln |P  150|=0.06t+C,PP150=±eCe0.06t=Be0.06t[where B=±eC].
  1. Substitution of t=0 and P=200 into (15) gives B=4. With this value of B we solve Eq. (15) for

    (16)P(t)=600e0.06t4e0.06t1.

    Note that, as t increases and approaches T=ln (4)/0.0623.105, the positive denominator on the right in (16) decreases and approaches 0. Consequently P(t)+ as tT. This is a doomsday situation—a real population explosion.

  2. Substitution of t=0 and P=100 into (15) gives B=2. With this value of B we solve Eq. (15) for

    (17)P(t)=300e0.06t2e0.06t+1=3002+e0.06t.

    Note that, as t increases without bound, the positive denominator on the right in (16) approaches +. Consequently, P(t)0 as t+. This is an (eventual) extinction situation.

Thus the population in Example 7 either explodes or is an endangered species threatened with extinction, depending on whether or not its initial size exceeds the threshold population M=150. An approximation to this phenomenon is sometimes observed with animal populations, such as the alligator population in certain areas of the southern United States.

Figure 2.1.6 shows typical solution curves that illustrate the two possibilities for a population P(t) satisfying Eq. (13). If P0=M (exactly!), then the population remains constant. However, this equilibrium situation is very unstable. If P0 exceeds M (even slightly), then P(t) rapidly increases without bound, whereas if the initial (positive) population is less than M (however slightly), then it decreases (more gradually) toward zero as t+. See Problem 33.

FIGURE 2.1.6.

Typical solution curves for the explosion/extinction equation P=kP(PM).

2.1 Problems

Separate variables and use partial fractions to solve the initial value problems in Problems 1–8. Use either the exact solution or a computer-generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution.

  1. dxdt=xx2, x(0)=2

  2. dxdt=10xx2, x(0)=1

  3. dxdt=1x2, x(0)=3

  4. dxdt=94x2, x(0)=0

  5. dxdt=3x(5x), x(0)=8

  6. dxdt=3x(x5), x(0)=2

  7. dxdt=4x(7x), x(0)=11

  8. dxdt=7x(x13), x(0)=17

  9. Population growth The time rate of change of a rabbit population P is proportional to the square root of P. At time t=0 (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many rabbits will there be one year later?

  10. Extinction by disease Suppose that the fish population P(t) in a lake is attacked by a disease at time t=0, with the result that the fish cease to reproduce (so that the birth rate is β=0) and the death rate δ (deaths per week per fish) is thereafter proportional to 1/P. If there were initially 900 fish in the lake and 441 were left after 6 weeks, how long did it take all the fish in the lake to die?

  11. Fish population Suppose that when a certain lake is stocked with fish, the birth and death rates β and δ are both inversely proportional to P. (a) Show that

    P(t)=(12kt+P0)2,

    where k is a constant. (b) If P0=100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?

  12. Population growth The time rate of change of an alligator population P in a swamp is proportional to the square of P. The swamp contained a dozen alligators in 1988, two dozen in 1998. When will there be four dozen alligators in the swamp? What happens thereafter?

  13. Birth rate exceeds death rate Consider a prolific breed of rabbits whose birth and death rates, β and δ, are each proportional to the rabbit population P=P(t), with β>δ. (a) Show that

    P(t)=P01kP0t,k constant.

    Note that P(t)+ as t1/(kP0). This is doomsday. (b) Suppose that P0=6 and that there are nine rabbits after ten months. When does doomsday occur?

  14. Death rate exceeds birth rate Repeat part (a) of Problem 13 in the case β<δ. What now happens to the rabbit population in the long run?

  15. Limiting population Consider a population P(t) satisfying the logistic equation dP/dt=aPbP2, where B=aP is the time rate at which births occur and D=bP2 is the rate at which deaths occur. If the initial population is P(0)=P0, and B0 births per month and D0 deaths per month are occurring at time t=0, show that the limiting population is M=B0P0/D0.

  16. Limiting population Consider a rabbit population P(t) satisfying the logistic equation as in Problem 15. If the initial population is 120 rabbits and there are 8 births per month and 6 deaths per month occurring at time t=0, how many months does it take for P(t) to reach 95% of the limiting population M?

  17. Limiting population Consider a rabbit population P(t) satisfying the logistic equation as in Problem 15. If the initial population is 240 rabbits and there are 9 births per month and 12 deaths per month occurring at time t=0, how many months does it take for P(t) to reach 105% of the limiting population M?

  18. Threshold population Consider a population P(t) satisfying the extinction-explosion equation dP/dt=aP2bP, where B=aP2 is the time rate at which births occur and D=bP is the rate at which deaths occur. If the initial population is P(0)=P0 and B0 births per month and D0 deaths per month are occurring at time t=0, show that the threshold population is M=D0P0/B0.

  19. Threshold population Consider an alligator population P(t) satisfying the extinction-explosion equation as in Problem 18. If the initial population is 100 alligators and there are 10 births per month and 9 deaths per month occurring at time t=0, how many months does it take for P(t) to reach 10 times the threshold population M?

  20. Threshold population Consider an alligator population P(t) satisfying the extinction-explosion equation as in Problem 18. If the initial population is 110 alligators and there are 11 births per month and 12 deaths per month occurring at time t=0, how many months does it take for P(t) to reach 10% of the threshold population M?

  21. Logistic model Suppose that the population P(t) of a country satisfies the differential equation dP/dt=kP(200P) with k constant. Its population in 1960 was 100 million and was then growing at the rate of 1 million per year. Predict this country’s population for the year 2020.

  22. Logistic model Suppose that at time t=0, half of a “logistic” population of 100,000 persons have heard a certain rumor, and that the number of those who have heard it is then increasing at the rate of 1000 persons per day. How long will it take for this rumor to spread to 80% of the population? (Suggestion: Find the value of k by substituting P(0) and P(0) in the logistic equation, Eq. (3).)

  23. Solution rate As the salt KNO3 dissolves in methanol, the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation dx/dt=0.8x0.004x2.

    1. What is the maximum amount of the salt that will ever dissolve in the methanol?

    2. If x=50 when t=0, how long will it take for an additional 50 g of salt to dissolve?

  24. Spread of disease Suppose that a community contains 15,000 people who are susceptible to Michaud’s syndrome, a contagious disease. At time t=0 the number N(t) of people who have developed Michaud’s syndrome is 5000 and is increasing at the rate of 500 per day. Assume that N(t) is proportional to the product of the numbers of those who have caught the disease and of those who have not. How long will it take for another 5000 people to develop Michaud’s syndrome?

  25. Logistic model The data in the table in Fig. 2.1.7 are given for a certain population P(t) that satisfies the logistic equation in (3). (a) What is the limiting population M? (Suggestion: Use the approximation

    P(t)P(t+h)P(th)2h

    with h=1 to estimate the values of P(t) when P=25.00 and when P=47.54. Then substitute these values in the logistic equation and solve for k and M.) (b) Use the values of k and M found in part (a) to determine when P=75. (Suggestion: Take t=0 to correspond to the year 1965.)

    FIGURE 2.1.7.

    Population data for Problem 25.

    Year P (millions)
    1964 24.63
    1965 25.00
    1966 25.38
    2014 47.04
    2015 47.54
    2016 48.04
  26. Constant death rate A population P(t) of small rodents has birth rate β=(0.001)P (births per month per rodent) and constant death rate δ. If P(0)=100 and P(0)=8, how long (in months) will it take this population to double to 200 rodents? (Suggestion: First find the value of δ.)

  27. Constant death rate Consider an animal population P(t) with constant death rate δ=0.01 (deaths per animal per month) and with birth rate β proportional to P. Suppose that P(0)=200 and P(0)=2. (a) When is P=1000? (b) When does doomsday occur?

  28. Population growth Suppose that the number x(t) (with t in months) of alligators in a swamp satisfies the differential equation dx/dt=0.0001x20.01x.

    1. If initially there are 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.

    2. Repeat part (a), except with 150 alligators initially.

  29. Logistic model During the period from 1790 to 1930, the U.S. population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this period, P(t) remained close to the solution of the initial value problem

    dPdt=0.03135P0.0001489P2,P(0)=3.9.
    1. What 1930 population does this logistic equation predict?

    2. What limiting population does it predict?

    3. Has this logistic equation continued since 1930 to accurately model the U.S. population?

    [This problem is based on a computation by Verhulst, who in 1845 used the 1790–1840 U.S. population data to predict accurately the U.S. population through the year 1930 (long after his own death, of course).]

  30. Tumor growth A tumor may be regarded as a population of multiplying cells. It is found empirically that the “birth rate” of the cells in a tumor decreases exponentially with time, so that β(t)=β0eαt (where α and β0 are positive constants), and hence

    dPdt=β0eαtP,P(0)=P0.

    Solve this initial value problem for

    P(t)=P0exp(β0α(1eαt)).

    Observe that P(t) approaches the finite limiting population P0exp(β0/α) as t+.

  31. Tumor growth For the tumor of Problem 30, suppose that at time t=0 there are P0=106 cells and that P(t) is then increasing at the rate of 3×105 cells per month. After 6 months the tumor has doubled (in size and in number of cells). Solve numerically for α, and then find the limiting population of the tumor.

  32. Derive the solution

    P(t)=MP0P0+(MP0)ekMt

    of the logistic initial value problem P=kP(MP), P(0)=P0. Make it clear how your derivation depends on whether 0<P0<M or P0>M.

    1. Derive the solution

      P(t)=MP0P0+(MP0)ekMt

      of the extinction-explosion initial value problem P=kP(PM), P(0)=P0.

    2. How does the behavior of P(t) as t increases depend on whether 0<P0<M or P0>M?

  33. If P(t) satisfies the logistic equation in (3), use the chain rule to show that

    P(t)=2k2P(P12M)(PM).

    Conclude that P>0 if 0<P<12M; P=0 if P=12M; P<0 if 12M<P<M; and P>0 if P>M. In particular, it follows that any solution curve that crosses the line P=12M has an inflection point where it crosses that line, and therefore resembles one of the lower S-shaped curves in Fig. 2.1.3.

  34. Approach to limiting population Consider two population functions P1(t) and P2(t), both of which satisfy the logistic equation with the same limiting population M but with different values k1 and k2 of the constant k in Eq. (3). Assume that k1<k2. Which population approaches M the most rapidly? You can reason geometrically by examining slope fields (especially if appropriate software is available), symbolically by analyzing the solution given in Eq. (7), or numerically by substituting successive values of t.

  35. Logistic modeling To solve the two equations in (10) for the values of k and M, begin by solving the first equation for the quantity x=e50kM and the second equation for x2=e100kM. Upon equating the two resulting expressions for x2 in terms of M, you get an equation that is readily solved for M. With M now known, either of the original equations is readily solved for k. This technique can be used to “fit” the logistic equation to any three population values P0, P1, and P2 corresponding to equally spaced times t0=0, t1, and t2=2t1.

  36. Logistic modeling Use the method of Problem 36 to fit the logistic equation to the actual U.S. population data (Fig. 2.1.4) for the years 1850, 1900, and 1950. Solve the resulting logistic equation and compare the predicted and actual populations for the years 1990 and 2000.

  37. Logistic modeling Fit the logistic equation to the actual U.S. population data (Fig. 2.1.4) for the years 1900, 1930, and 1960. Solve the resulting logistic equation, then compare the predicted and actual populations for the years 1980, 1990, and 2000.

  38. Periodic growth rate Birth and death rates of animal populations typically are not constant; instead, they vary periodically with the passage of seasons. Find P(t) if the population P satisfies the differential equation

    dPdt=(k+b cos 2πt)P,

    where t is in years and k and b are positive constants. Thus the growth-rate function r(t)=k+bcos 2πt varies periodically about its mean value k. Construct a graph that contrasts the growth of this population with one that has the same initial value P0 but satisfies the natural growth equation P=kP (same constant k). How would the two populations compare after the passage of many years?

2.1 Application Logistic Modeling of Population Data

These investigations deal with the problem of fitting a logistic model to given population data. Thus we want to determine the numerical constants a and b so that the solution P(t) of the initial value problem

(1)dPdt=aP+bP2,P(0)=P0

approximates the given values P0, P1, , Pn of the population at the times t0=0, t1, , tn. If we rewrite Eq. (1) (the logistic equation with kM=a and k=b) in the form

(2)1PdPdt=a+bP,

then we see that the points

(P(ti), P(ti)P(ti)),i=0, 1, 2, , n,

should all lie on the straight line with y-intercept a and slope b (as determined by the function of P on the right-hand side in Eq. (2)).

This observation provides a way to find a and b. If we can determine the approximate values of the derivatives P1,P2, corresponding to the given population data, then we can proceed with the following agenda:

  • First plot the points (P1,P1/P1), (P2,P2/P2),  on a sheet of graph paper with horizontal P-axis.

  • Then use a ruler to draw a straight line that appears to approximate these points well.

  • Finally, measure this straight line’s y-intercept a and slope b.

But where are we to find the needed values of the derivative P(t) of the (as yet) unknown function P? It is easiest to use the approximation

(3)Pi=Pi + 1Pi  1ti + 1ti  1

suggested by Fig. 2.1.8. For instance, if we take i=0 corresponding to the year 1790, then the U.S. population data in Fig. 2.1.9 give

P1=P2P0t2t0=7.2403.929200.166

for the slope at (t1,P1) corresponding to the year 1800.

FIGURE 2.1.8.

The symmetric difference approximation Pi + 1Pi  1ti + 1ti  1 to the derivative P(ti).

Investigation A: Use Eq. (3) to verify the slope figures shown in the final column of the table in Fig. 2.1.9, then plot the points (P1,P1/P1), , (P11,P11/P11) indicated by the dots in Fig. 2.1.10. If an appropriate graphing calculator, spreadsheet, or computer program is available, use it to find the straight line y=a+bP as in (2) that best fits these points. If not, draw your own straight line approximating these points, and then measure its intercept a and slope b as accurately as you can. Next, solve the logistic equation in (1) with these numerical parameters, taking t=0 corresponding to the year 1800. Finally, compare the predicted 20th- and 21st-century U.S. population figures with the actual data listed in Fig. 2.1.4.

FIGURE 2.1.9.

U.S. population data (in millions) and approximate slopes.

Year i ti Population Pi Slope Pi
1790 0 10 3.929  
1800 1 0 5.308 0.166
1810 2 10 7.240 0.217
1820 3 20 9.638 0.281
1830 4 30 12.861 0.371
1840 5 40 17.064 0.517
1850 6 50 23.192 0.719
1860 7 60 31.443 0.768
1870 8 70 38.558 0.937
1880 9 80 50.189 1.221
1890 10 90 62.980 1.301
1900 11 100 76.212 1.462
1910 12 110 92.228  

FIGURE 2.1.10.

Points and approximating straight line for U.S. population data from 1800 to 1900.

FIGURE 2.1.11.

World population data.

Year World Population (billions)
1975 4.062
1980 4.440
1985 4.853
1990 5.310
1995 5.735
2000 6.127
2005 6.520
2010 6.930
2015 7.349

Investigation B: Repeat Investigation A, but take t=0 in 1900 and use only the 20th-century population data listed in Fig. 2.1.4 to create a logistic model. How well does your model predict the U.S. population in the years 1990–2010?

Investigation C: Model similarly the world population data shown in Fig. 2.1.11. The Population Division of the United Nations predicts a world population of 9.157 billion in the year 2040. What do you predict?

2.2 Equilibrium Solutions and Stability

In previous sections we have often used explicit solutions of differential equations to answer specific numerical questions. But even when a given differential equation is difficult or impossible to solve explicitly, it often is possible to extract qualitative information about general properties of its solutions. For example, we may be able to establish that every solution x(t) grows without bound as t+, or approaches a finite limit, or is a periodic function of t. In this section we introduce—mainly by consideration of simple differential equations that can be solved explicitly—some of the more important qualitative questions that can sometimes be answered for equations that are difficult or impossible to solve.

Example 1

Cooling and heating Let x(t) denote the temperature of a body with initial temperature x(0)=x0. At time t=0 this body is immersed in a medium with constant temperature A. Assuming Newton’s law of cooling,

(1)dxdt=k(xA)(k>0constant),

we readily solve (by separation of variables) for the explicit solution

x(t)=A+(x0A)ekt.

It follows immediately that

(2)limtx(t)=A,

so the temperature of the body approaches that of the surrounding medium (as is evident to one’s intuition). Note that the constant function x(t)A is a solution of Eq. (1); it corresponds to the temperature of the body when it is in thermal equilibrium with the surrounding medium. In Fig. 2.2.1 the limit in (2) means that every other solution curve approaches the equilibrium solution curve x=A asymptotically as t+.

FIGURE 2.2.1.

Typical solution curves for the equation of Newton’s law of cooling, dx/dt=k(xA).

Remark

The behavior of solutions of Eq. (1) is summarized briefly by the phase diagram in Fig. 2.2.2—which indicates the direction (or “phase”) of change in x as a function of x itself. The right-hand side f(x)=k(xA)=k(Ax) is positive if x<A, negative if x>A. This observation corresponds to the fact that solutions starting above the line x=A and those starting below it both approach the limiting solution x(t)A as t increases (as indicated by the arrows).

FIGURE 2.2.2.

Phase diagram for the equation dx/dt=f(x)=k(Ax).

In Section 2.1 we introduced the general population equation

(3)dxdt=(βδ)x,

where β and δ are the birth and death rates, respectively, in births or deaths per individual per unit of time. The question of whether a population x(t) is bounded or unbounded as t+ is of evident interest. In many situations—like the logistic and explosion/extinction populations of Section 2.1—the birth and death rates are known functions of x. Then Eq. (3) takes the form

(4)dxdt=f(x).

This is an autonomous first-order differential equation—one in which the independent variable t does not appear explicitly (the terminology here stemming from the Greek word autonomos for “independent,” e.g., of the time t). As in Example 1, the solutions of the equation f(x)=0 play an important role and are called critical points of the autonomous differential equation dx/dt=f(x).

If x=c is a critical point of Eq. (4), then the differential equation has the constant solution x(t)c. A constant solution of a differential equation is sometimes called an equilibrium solution (one may think of a population that remains constant because it is in “equilibrium” with its environment). Thus the critical point x=c, a number, corresponds to the equilibrium solution x(t)c, a constant-valued function.

Example 2 illustrates the fact that the qualitative behavior (as t increases) of the solutions of an autonomous first-order equation can be described in terms of its critical points.

Example 2

Logistic equation Consider the logistic differential equation

(5)dxdt=kx(Mx)

(with k>0 and M>0). It has two critical points—the solutions x=0 and x=M of the equation

f(x)=kx(Mx)=0.

In Section 2.1 we discussed the logistic-equation solution

(6)x(t)=Mx0x0+(Mx0)ekMt

satisfying the initial condition x(0)=x0. Note that the initial values x0=0 and x0=M yield the equilibrium solutions x(t)0 and x(t)M of Eq. (5).

We observed in Section 2.1 that if x0>0, then x(t)M as t+. But if x0<0, then the denominator in Eq. (6) initially is positive, but vanishes when

t=t1=1kMlnMx0x0>0.

Because the numerator in (6) is negative in this case, it follows that

limtt1x(t)=ifx0<0.

It follows that the solution curves of the logistic equation in (5) look as illustrated in Fig. 2.2.3. Here we see graphically that every solution either approaches the equilibrium solution x(t)M as t increases, or (in a visually obvious sense) diverges away from the other equilibrium solution x(t)0.

Stability of Critical Points

Figure 2.2.3 illustrates the concept of stability. A critical point x=c of an autonomous first-order equation is said to be stable provided that, if the initial value x0 is sufficiently close to c, then x(t) remains close to c for all t>0. More precisely, the critical point c is stable if, for each ϵ>0, there exists δ>0 such that

(7)|x0c|<δimplies that|x(t)c|<ϵ

for all t>0. The critical point x=c is unstable if it is not stable.

FIGURE 2.2.3.

Typical solution curves for the logistic equation dx/dt=kx(Mx).

Figure 2.2.4 shows a “wider view” of the solution curves of a logistic equation with k=1 and M=4. Note that the strip 3.5<x<4.5 enclosing the stable equilibrium curve x=4 acts like a funnel—solution curves (moving from left to right) enter this strip and thereafter remain within it. By contrast, the strip 0.5<x<0.5 enclosing the unstable solution curve x=0 acts like a spout—solution curves leave this strip and thereafter remain outside it. Thus the critical point x=M is stable, whereas the critical point x=0 is unstable.

FIGURE 2.2.4.

Solution curves, funnel, and spout for dx/dt=4xx2.

Remark 1

We can summarize the behavior of solutions of the logistic equation in (5)—in terms of their initial values—by means of the phase diagram shown in Fig. 2.2.5. It indicates that x(t)M as t+ if either x0>M or 0<x0<M, whereas x(t) as t increases if x0<0. The fact that M is a stable critical point would be important, for instance, if we wished to conduct an experiment with a population of M bacteria. It is impossible to count precisely M bacteria for M large, but any initially positive population will approach M as t increases.

FIGURE 2.2.5.

Phase diagram for the logistic equation dx/dt=f(x)=kx(Mx).

Remark 2

Related to the stability of the limiting solution M=a/b of the logistic equation

(8)dxdt=axbx2

is the “predictability” of M for an actual population. The coefficients a and b are unlikely to be known precisely for an actual population. But if they are replaced with close approximations a and b—derived perhaps from empirical measurements—then the approximate limiting population M=a/b will be close to the actual limiting population M=a/b. We may therefore say that the value M of the limiting population predicted by a logistic equation not only is a stable critical point of the differential equation, but this value also is “stable” with respect to small perturbations of the constant coefficients in the equation. (Note that one of these two statements involves changes in the initial value x0; the other involves changes in the coefficients a and b.)

Example 3

Explosion/extinction Consider now the explosion/extinction equation

(9)dxdt=kx(xM)

of Eq. (10) in Section 2.1. Like the logistic equation, it has the two critical points x=0 and x=M corresponding to the equilibrium solutions x(t)0 and x(t)M. According to Problem 33 in Section 2.1, its solution with x(0)=x0 is given by

(10)x(t)=Mx0x0+(Mx0)ekMt

(with only a single difference in sign from the logistic solution in (6)). If x0<M, then (because the coefficient of the exponential in the denominator is positive) it follows immediately from Eq. (10) that x(t)0 as t+. But if x0>M, then the denominator in (10) initially is positive, but vanishes when

t=t1=1kMlnx0x0M>0.

Because the numerator in (10) is positive in this case, it follows that

limtt1x(t)=+ifx0>M.

Therefore, the solution curves of the explosion/extinction equation in (9) look as illustrated in Fig. 2.2.6. A narrow band along the equilibrium curve x=0 (as in Fig. 2.2.4) would serve as a funnel, while a band along the solution curve x=M would serve as a spout for solutions. The behavior of the solutions of Eq. (9) is summarized by the phase diagram in Fig. 2.2.7, where we see that the critical point x=0 is stable and the critical point x=M is unstable.

FIGURE 2.2.6.

Typical solution curves for the explosion/extinction equation dx/dt=kx(xM).

FIGURE 2.2.7.

Phase diagram for the explosion/extinction equation dx/dt=f(x)=kx(xM).

Harvesting a Logistic Population

The autonomous differential equation

(11)dxdt=axbx2h

(with a, b, and h all positive) may be considered to describe a logistic population with harvesting. For instance, we might think of the population of fish in a lake from which h fish per year are removed by fishing.

Example 4

Threshold/limiting population Let us rewrite Eq. (11) in the form

(12)dxdt=kx(Mx)h,

which exhibits the limiting population M in the case h=0 of no harvesting. Assuming hereafter that h>0, we can solve the quadratic equation kx2+kMxh=0 for the two critical points

(13)H, N=kM±(kM)24hk2k=12(M±M24h/k),

assuming that the harvesting rate h is sufficiently small that 4h<kM2, so both roots H and N are real with 0<H<N<M. Then we can rewrite Eq. (12) in the form

(14)dxdt=k(Nx)(xH).

For instance, the number of critical points of the equation may change abruptly as the value of a parameter is changed. In Problem 24 we ask you to solve this equation for the solution

(15)x(t)=N(x0H)H(x0N)ek(NH)t(x0H)(x0N)ek(NH)t

in terms of the initial value x(0)=x0.

Note that the exponent k(NH)t is negative for t>0. If x0>N, then each of the coefficients within parentheses in Eq. (15) is positive; it follows that

(16)Ifx0>Nthenx(t)Nast+.

In Problem 25 we ask you to deduce also from Eq. (15) that

(17)IfH<x0<Nthenx(t)Nast+,whereas
(18)ifx0<Hthenx(t)astt1

for a positive value t1 that depends on x0. It follows that the solution curves of Eq. (12)—still assuming that 4h<kM2—look as illustrated in Fig. 2.2.8. (Can you visualize a funnel along the line x=N and a spout along the line x=H?) Thus the constant solution x(t)N is an equilibrium limiting solution, whereas x(t)H is a threshold solution that separates different behaviors—the population approaches N if x0>H, while it becomes extinct because of harvesting if x0<H. Finally, the stable critical point x=N and the unstable critical point x=H are illustrated in the phase diagram in Fig. 2.2.9.

FIGURE 2.2.8.

Typical solution curves for the logistic harvesting equation dx/dt=k(Nx)(xH).

FIGURE 2.2.9.

Phase diagram for the logistic harvesting equation dx/dt=f(x)=k(Nx)(xH).

Example 5

Lake stocked with fish For a concrete application of our stability conclusions in Example 4, suppose that k=1 and M=4 for a logistic population x(t) of fish in a lake, measured in hundreds after t years. Without any fishing at all, the lake would eventually contain nearly 400 fish, whatever the initial population. Now suppose that h=3, so that 300 fish are “harvested” annually (at a constant rate throughout the year). Equation (12) is then dx/dt=x(4x)3, and the quadratic equation

x2+4x3=(3x)(x1)=0

has solutions H=1 and N=3. Thus the threshold population is 100 fish and the (new) limiting population is 300 fish. In short, if the lake is stocked initially with more than 100 fish, then as t increases, the fish population will approach a limiting value of 300 fish. But if the lake is stocked initially with fewer than 100 fish, then the lake will be “fished out” and the fish will disappear entirely within a finite period of time.

Bifurcation and Dependence on Parameters

A biological or physical system that is modeled by a differential equation may depend crucially on the numerical values of certain coefficients or parameters that appear in the equation. For instance, the number of critical points of the equation may change abruptly as the value of a parameter is changed.

Example 6

Critical/excessive harvesting The differential equation

(19)dxdt=x(4x)h

(with x in hundreds) models the harvesting of a logistic population with k=1 and limiting population M=4 (hundred). In Example 5 we considered the case of harvesting level h=3, and found that the new limiting population is N=3 hundred and the threshold population is H=1 hundred. Typical solution curves, including the equilibrium solutions x(t)3 and x(t)1, then look like those pictured in Fig. 2.2.8.

Now let’s investigate the dependence of this picture upon the harvesting level h. According to Eq. (13) with k=1 and M=4, the limiting and threshold populations N and H are given by

(20)H, N=12(4±164h)=2±4h.

If h<4—we can consider negative values of h to describe stocking rather than harvesting the fish—then there are distinct equilibrium solutions x(t)N and x(t)H with N>H as in Fig. 2.2.8.

But if h=4, then Eq. (20) gives N=H=2, so the differential equation has only the single equilibrium solution x(t)2. In this case the solution curves of the equation look like those illustrated in Fig. 2.2.10. If the initial number x0 (in hundreds) of fish exceeds 2, then the population approaches a limiting population of 2 (hundred fish). However, any initial population x0<2 (hundred) results in extinction with the fish dying out as a consequence of the harvesting of 4 hundred fish annually. The critical point x=2 might therefore be described as “semistable”—it looks stable on the side x>2 where solution curves approach the equilibrium solution x(t)2 as t increases, but unstable on the side x<2 where solution curves instead diverge away from the equilibrium solution.

FIGURE 2.2.10.

Solution curves of the equation x=x(4x)h with critical harvesting h=4.

If, finally, h>4, then the quadratic equation corresponding to (20) has no real solutions and the differential equation in (19) has no equilibrium solutions. The solution curves then look like those illustrated in Fig. 2.2.11, and (whatever the initial number of fish) the population dies out as a result of the excessive harvesting.

FIGURE 2.2.11.

Solution curves of the equation x=x(4x)h with excessive harvesting h=5.

If we imagine turning a dial to gradually increase the value of the parameter h in Eq. (19), then the picture of the solution curves changes from one like Fig. 2.2.8 with h<4, to Fig. 2.2.10 with h=4, to one like Fig. 2.2.11 with h>4. Thus the differential equation has

The value h=4—for which the qualitative nature of the solutions changes as h increases—is called a bifurcation point for the differential equation containing the parameter h. A common way to visualize the corresponding “bifurcation” in the solutions is to plot the bifurcation diagram consisting of all points (h, c), where c is a critical point of the equation x=x(4x)+h. For instance, if we rewrite Eq. (20) as

c=2±4h,(c2)2=4h,

where either c=N or c=H, then we get the equation of the parabola that is shown in Fig. 2.2.12. This parabola is then the bifurcation diagram for our differential equation that models a logistic fish population with harvesting at the level specified by the parameter h.

FIGURE 2.2.12.

The parabola (c2)2=4h is the bifurcation diagram for the differential equation x=x(4x)h.

2.2 Problems

In Problems 1 through 12 first solve the equation f(x)=0 to find the critical points of the given autonomous differential equation dx/dt=f(x). Then analyze the sign of f(x) to determine whether each critical point is stable or unstable, and construct the corresponding phase diagram for the differential equation. Next, solve the differential equation explicitly for x(t) in terms of t. Finally, use either the exact solution or a computer-generated slope field to sketch typical solution curves for the given differential equation, and verify visually the stability of each critical point.

  1. dxdt=x4

  2. dxdt=3x

  3. dxdt=x24x

  4. dxdt=3xx2

  5. dxdt=x24

  6. dxdt=9x2

  7. dxdt=(x2)2

  8. dxdt=(3x)2

  9. dxdt=x25x+4

  10. dxdt=7xx210

  11. dxdt=(x1)3

  12. dxdt=(2x)3

In Problems 13 through 18, use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate the stability or instability of each critical point of the given differential equation. (Some of these critical points may be semistable in the sense mentioned in Example 6.)

  1. dxdt=(x+2)(x2)2

  2. dxdt=x(x24)

  3. dxdt=(x24)2

  4. dxdt=(x24)3

  5. dxdt=x2(x24)

  6. dxdt=x3(x24)

  7. The differential equation dx/dt=110x(10x)h models a logistic population with harvesting at rate h. Determine (as in Example 6) the dependence of the number of critical points on the parameter h, and then construct a bifurcation diagram like Fig. 2.2.12.

  8. The differential equation dx/dt=1100x(x5)+s models a population with stocking at rate s. Determine the dependence of the number of critical points c on the parameter s, and then construct the corresponding bifurcation diagram in the sc-plane.

  9. Pitchfork bifurcation Consider the differential equation dx/dt=kxx3. (a) If k0, show that the only critical value c=0 of x is stable. (b) If k>0, show that the critical point c=0 is now unstable, but that the critical points c=±k are stable. Thus the qualitative nature of the solutions changes at k=0 as the parameter k increases, and so k=0 is a bifurcation point for the differential equation with parameter k. The plot of all points of the form (k, c) where c is a critical point of the equation x=kxx3 is the “pitchfork diagram” shown in Fig. 2.2.13.

    FIGURE 2.2.13.

    Bifurcation diagram for dx/dt=kxx3.

  10. Consider the differential equation dx/dt=x+kx3 containing the parameter k. Analyze (as in Problem 21) the dependence of the number and nature of the critical points on the value of k, and construct the corresponding bifurcation diagram.

  11. Variable-rate harvesting Suppose that the logistic equation dx/dt=kx(Mx) models a population x(t) of fish in a lake after t months during which no fishing occurs. Now suppose that, because of fishing, fish are removed from the lake at the rate of hx fish per month (with h a positive constant). Thus fish are “harvested” at a rate proportional to the existing fish population, rather than at the constant rate of Example 4. (a) If 0<h<kM, show that the population is still logistic. What is the new limiting population? (b) If hkM, show that x(t)0 are t+, so the lake is eventually fished out.

  12. Separate variables in the logistic harvesting equation dx/dt=k(Nx)(xH) and then use partial fractions to derive the solution given in Eq. (15).

  13. Use the alternative forms

    x(t)=N(x0  H) + H(N  x0)e  k(N  H)t(x0  H) + (N  x0)e  k(N  H)t=H(N  x0)e  k(N  H)t  N(H  x0)(N  x0)e  k(N  H)t  (H  x0)

    of the solution in (15) to establish the conclusions stated in (17) and (18).

Constant-Rate Harvesting

Example 4 dealt with the case 4h>kM2 in the equation dx/dt=kx(Mx)h that describes constant-rate harvesting of a logistic population. Problems 26 and 27 deal with the other cases.

  1. If 4h=kM2, show that typical solution curves look as illustrated in Fig. 2.2.14. Thus if x0M/2, then x(t)M/2 as t+. But if x0<M/2, then x(t)=0 after a finite period of time, so the lake is fished out. The critical point x=M/2 might be called semistable, because it looks stable from one side, unstable from the other.

    FIGURE 2.2.14.

    Solution curves for harvesting a logistic population with 4h=kM2.

  2. If 4h>kM2, show that x(t)=0 after a finite period of time, so the lake is fished out (whatever the initial population). [Suggestion: Complete the square to rewrite the differential equation in the form dx/dt=k[(xa)2+b2]. Then solve explicitly by separation of variables.] The results of this and the previous problem (together with Example 4) show that h=14kM2 is a critical harvesting rate for a logistic population. At any lesser harvesting rate the population approaches a limiting population N that is less than M (why?), whereas at any greater harvesting rate the population reaches extinction.

  3. Alligator population This problem deals with the differential equation dx/dt=kx(xM)h that models the harvesting of an unsophisticated population (such as alligators). Show that this equation can be rewritten in the form dx/dt=k(xH)(xK), where

    H=12(M+M2+4h/k)>0,K=12(MM2+4h/k)<0.

    Show that typical solution curves look as illustrated in Fig. 2.2.15.

  4. Consider the two differential equations

    (21)dxdt=(xa)(xb)(xc)
    (22)anddxdt=(ax)(bx)(cx),

    each having the critical points a, b, and c; suppose that a<b<c. For one of these equations, only the critical point b is stable; for the other equation, b is the only unstable critical point. Construct phase diagrams for the two equations to determine which is which. Without attempting to solve either equation explicitly, make rough sketches of typical solution curves for each. You should see two funnels and a spout in one case, two spouts and a funnel in the other.

    FIGURE 2.2.15.

    Solution curves for harvesting a population of alligators.

2.3 Acceleration–Velocity Models

In Section 1.2 we discussed vertical motion of a mass m near the surface of the earth under the influence of constant gravitational acceleration. If we neglect any effects of air resistance, then Newton’s second law (F=ma) implies that the velocity v of the mass m satisfies the equation

(1)mdvdt=FG,

where FG=mg is the (downward-directed) force of gravity, where the gravitational acceleration is g9.8 m/s2 (in mks units; g32 ft/s2 in fps units).

Example 1

No air resistance Suppose that a crossbow bolt is shot straight upward from the ground (y0=0) with initial velocity v0=49 (m/s). Then Eq. (1) with g=9.8 gives

dvdt=9.8,sov(t)=(9.8)t+v0=(9.8)t+49.

Hence the bolt’s height function y(t) is given by

y(t)=[(9.8)t+49]dt=(4.9)t2+49t+y0=(4.9)t2+49t.

The bolt reaches its maximum height when v=(9.8)t+49=0, hence when t=5 (s). Thus its maximum height is

ymax=y(5)=(4.9)(52)+(49)(5)=122.5(m).

The bolt returns to the ground when y=(4.9)t(t10)=0, and thus after 10 seconds aloft.

Now we want to take account of air resistance in a problem like Example 1. The force FR exerted by air resistance on the moving mass m must be added in Eq. (1), so now

(2)mdvdt=FG+FR.

Newton showed in his Principia Mathematica that certain simple physical assumptions imply that FR is proportional to the square of the velocity: FR=kv2. But empirical investigations indicate that the actual dependence of air resistance on velocity can be quite complicated. For many purposes it suffices to assume that

FR=kvp,

where 1p2 and the value of k depends on the size and shape of the body, as well as the density and viscosity of the air. Generally speaking, p=1 for relatively low speeds and p=2 for high speeds, whereas 1<p<2 for intermediate speeds. But how slow “low speed” and how fast “high speed” are depend on the same factors that determine the value of the coefficient k.

Thus air resistance is a complicated physical phenomenon. But the simplifying assumption that FR is exactly of the form given here, with either p=1 or p=2, yields a tractable mathematical model that exhibits the most important qualitative features of motion with resistance.

Resistance Proportional to Velocity

Let us first consider the vertical motion of a body with mass m near the surface of the earth, subject to two forces: a downward gravitational force FG and a force FR of air resistance that is proportional to velocity (so that p=1) and of course directed opposite the direction of motion of the body. If we set up a coordinate system with the positive y-direction upward and with y=0 at ground level, then FG=mg and

(3)FR=kv,

where k is a positive constant and v=dy/dt is the velocity of the body. Note that the minus sign in Eq. (3) makes FR positive (an upward force) if the body is falling (v is negative) and makes FR negative (a downward force) if the body is rising (v is positive). As indicated in Fig. 2.3.1, the net force acting on the body is then

F=FR+FG=kvmg,

FIGURE 2.3.1.

Vertical motion with air resistance.

and Newton’s law of motion F=m(dv/dt) yields the equation

mdvdt=kvmg.

Thus

(4)dvdt=ρvg,

where ρ=k/m>0. You should verify for yourself that if the positive y-axis were directed downward, then Eq. (4) would take the form dv/dt=ρv+g.

Equation (4) is a separable first-order differential equation, and its solution is

(5)v(t)=(v0+gρ)eρtgρ.

Here, v0=v(0) is the initial velocity of the body. Note that

(6)vτ=limtv(t)=gρ.

Thus the speed of a body falling with air resistance does not increase indefinitely; instead, it approaches a finite limiting speed, or terminal speed,

(7)|vτ|=gρ=mgk.

This fact is what makes a parachute a practical invention; it even helps explain the occasional survival of people who fall without parachutes from high-flying airplanes.

We now rewrite Eq. (5) in the form

(8)dydt=(v0vτ)eρt+vτ.

Integration gives

y(t)=1ρ(v0vτ)eρt+vτt+C.

We substitute 0 for t and let y0=y(0) denote the initial height of the body. Thus we find that C=y0+(v0vτ)/ρ, and so

(9)y(t)=y0+vτt+1ρ(v0vτ)(1eρt).

Equations (8) and (9) give the velocity v and height y of a body moving vertically under the influence of gravity and air resistance. The formulas depend on the initial height y0 of the body, its initial velocity v0, and the drag coefficient ρ, the constant such that the acceleration due to air resistance is aR=ρv. The two equations also involve the terminal velocity vτ defined in Eq. (6).

For a person descending with the aid of a parachute, a typical value of ρ is 1.5, which corresponds to a terminal speed of |vτ|21.3 ft/s, or about 14.5 mi/h. With an unbuttoned overcoat flapping in the wind in place of a parachute, an unlucky skydiver might increase ρ to perhaps as much as 0.5, which gives a terminal speed of |vτ|65 ft/s, about 44 mi/h. See Problems 10 and 11 for some parachute-jump computations.

Example 2

Velocity-proportional resistance We again consider a bolt shot straight upward with initial velocity v0=49 m/s from a crossbow at ground level. But now we take air resistance into account, with ρ=0.04 in Eq. (4). We ask how the resulting maximum height and time aloft compare with the values found in Example 1.

Solution

We substitute y0=0, v0=49, and vτ=g/ρ=245 in Eqs. (5) and (9), and obtain

v(t)=294et/25245,y(t)=7350245t7350et/25.

To find the time required for the bolt to reach its maximum height (when v=0), we solve the equation

v(t)=294et/25245=0

for tm=25ln (294/245)4.558 (s). Its maximum height is then ymax=v(tm)108.280 meters (as opposed to 122.5 meters without air resistance). To find when the bolt strikes the ground, we must solve the equation

y(t)=7350245t7350et/25=0.

Using Newton’s method, we can begin with the initial guess t0=10 and carry out the iteration tn+1=tny(tn)/y(tn) to generate successive approximations to the root. Or we can simply use the Solve command on a calculator or computer. We find that the bolt is in the air for tf9.411 seconds (as opposed to 10 seconds without air resistance). It hits the ground with a reduced speed of |v(tf)|43.227 m/s (as opposed to its initial velocity of 49 m/s).

Thus the effect of air resistance is to decrease the bolt’s maximum height, the total time spent aloft, and its final impact speed. Note also that the bolt now spends more time in descent (tftm4.853 s) than in ascent (tm4.558 s).

Resistance Proportional to Square of Velocity

Now we assume that the force of air resistance is proportional to the square of the velocity:

(10)FR=±kv2,

with k>0. The choice of signs here depends on the direction of motion, which the force of resistance always opposes. Taking the positive y-direction as upward, FR<0 for upward motion (when v>0) while FR>0 for downward motion (when v<0). Thus the sign of FR is always opposite that of v, so we can rewrite Eq. (10) as

(10′)FR=kv|v|.

Then Newton’s second law gives

mdvdt=FG+FR=mgkv|v|;

that is,

(11)dvdt=gρv|v|,

where ρ=k/m>0. We must discuss the cases of upward and downward motion separately.

Upward Motion: Suppose that a projectile is launched straight upward from the initial position y0 with initial velocity v0>0. Then Eq. (11) with v>0 gives the differential equation

(12)dvdt=gρv2=g(1+ρgv2).

In Problem 13 we ask you to make the substitution u=vρ/g and apply the familiar integral

11+u2du=tan1u+C

to derive the projectile’s velocity function

(13)v(t)=gρtan(C1tρg)withC1=tan1(v0ρg).

Because tanu du=ln |cos u|+C, a second integration (see Problem 14) yields the position function

(14)y(t)=y0+1ρln|cos(C1tρg)cosC1|.

Downward Motion: Suppose that a projectile is launched (or dropped) straight downward from the initial position y0 with initial velocity v00. Then Eq. (11) with v<0 gives the differential equation

(15)dvdt=g+ρv2=g(1ρgv2).

In Problem 15 we ask you to make the substitution u=vρ/g and apply the integral

11u2du=tanh1u+C

to derive the projectile’s velocity function

(16)v(t)=gρtanh(C2tρg)withC2=tanh1(v0ρg).

Because tanh u du=ln |coshu|+C, another integration (Problem 16) yields the position function

(17)y(t)=y01ρln|cosh(C2tρg)coshC2|.

(Note the analogy between Eqs. (16) and (17) and Eqs. (13) and (14) for upward motion.)

If v0=0, then C2=0, so v(t)=g/ρtanh(tρg). Because

limxtanhx=limxsinhxcoshx=limx12(exex)12(ex+ex)=1,

it follows that in the case of downward motion the body approaches the terminal speed

(18)|vτ|=gρ

(as compared with |vτ|=g/ρ in the case of downward motion with linear resistance described by Eq. (4)).

Example 3

Square-proportional resistance We consider once more a bolt shot straight upward with initial velocity v0=49 m/s from a crossbow at ground level, as in Example 2. But now we assume air resistance proportional to the square of the velocity, with ρ=0.0011 in Eqs. (12) and (15). In Problems 17 and 18 we ask you to verify the entries in the last line of the following table.

Air Resistance Maximum Height (ft) Time Aloft (s) Ascent Time (s) Descent Time (s) Impact Speed (ft/s)
0.0 122.5 10 5 5 49
(0.04)v 108.28 9.41 4.56 4.85 43.23
(0.0011)v2 108.47 9.41 4.61 4.80 43.49

Comparison of the last two lines of data here indicates little difference—for the motion of our crossbow bolt—between linear air resistance and air resistance proportional to the square of the velocity. And in Fig. 2.3.2, where the corresponding height functions are graphed, the difference is hardly visible. However, the difference between linear and nonlinear resistance can be significant in more complex situations—such as, for instance, the atmospheric reentry and descent of a space vehicle.

FIGURE 2.3.2.

The height functions in Example 1 (without air resistance), Example 2 (with linear air resistance), and Example 3 (with air resistance proportional to the square of the velocity) are all plotted. The graphs of the latter two are visually indistinguishable.

Variable Gravitational Acceleration

Unless a projectile in vertical motion remains in the immediate vicinity of the earth’s surface, the gravitational acceleration acting on it is not constant. According to Newton’s law of gravitation, the gravitational force of attraction between two point masses M and m located at a distance r apart is given by

(19)F=GMmr2,

where G is a certain empirical constant (G6.6726×1011 N·(m/kg)2 in mks units). The formula is also valid if either or both of the two masses are homogeneous spheres; in this case, the distance r is measured between the centers of the spheres.

The following example is similar to Example 2 in Section 1.2, but now we take account of lunar gravity.

Example 4

Lunar lander A lunar lander is free-falling toward the moon, and at an altitude of 53 kilometers above the lunar surface its downward velocity is measured at 1477 km/h. Its retrorockets, when fired in free space, provide a deceleration of T=4 m/s2. At what height above the lunar surface should the retrorockets be activated to ensure a “soft touchdown” (v=0 at impact)?

Solution

Let r(t) denote the lander’s distance from the center of the moon at time t (Fig. 2.3.3). When we combine the (constant) thrust acceleration T and the (negative) lunar acceleration F/m=GM/r2 of Eq. (19), we get the (acceleration) differential equation

(20)d2rdt2=TGMr2,

FIGURE 2.3.3.

The lunar lander descending to the surface of the moon.

where M=7.35×1022 (kg) is the mass of the moon, which has a radius of R=1.74×106 meters (or 1740 km, a little over a quarter of the earth’s radius). Noting that this second-order differential equation does not involve the independent variable t, we substitute

v=drdt,d2rdt2=dvdt=dvdrdrdt=vdvdr

(as in Eq. (36) of Section 1.6) and obtain the first-order equation

vdvdr=TGMr2

with the new independent variable r. Integration with respect to r now yields the equation

(21)12v2=Tr+GMr+C

that we can apply both before ignition (T=0) and after ignition (T=4).

Before ignition: Substitution of T=0 in (21) gives the equation

(21a)12v2=GMr+C1

where the constant is given by C1=v02/2GM/r0 with

v0=1477kmh×1000mkm×1 h3600 s=1477036ms

and r0=(1.74×106)+53,000=1.793×106 m (from the initial velocity–position measurement).

After ignition: Substitution of T=4 and v=0, r=R (at touchdown) into (21) gives

(21b)12v2=4r+GMr+C2

where the constant C2=4RGM/R is obtained by substituting the values v=0, r=R at touchdown.

At the instant of ignition the lunar lander’s position and velocity satisfy both (21a) and (21b). Therefore we can find its desired height h above the lunar surface at ignition by equating the right-hand sides in (21a) and (21b). This gives r=14(C1C2)=1.78187×106 and finally h=rR=41,870 meters (that is, 41.87 kilometers—just over 26 miles). Moreover, substitution of this value of r in (21a) gives the velocity v=450 m/s at the instant of ignition.

Escape Velocity

In his novel From the Earth to the Moon (1865), Jules Verne raised the question of the initial velocity necessary for a projectile fired from the surface of the earth to reach the moon. Similarly, we can ask what initial velocity v0 is necessary for the projectile to escape from the earth altogether. This will be so if its velocity v=dr/dt remains positive for all t>0, so it continues forever to move away from the earth. With r(t) denoting the projectile’s distance from the earth’s center at time t (Fig. 2.3.4), we have the equation

(22)dvdt=d2rdt2=GMr2,

FIGURE 2.3.4.

A mass m at a great distance from the earth.

similar to Eq. (20), but with T=0 (no thrust) and with M=5.975×1024 (kg) denoting the mass of the earth, which has an equatorial radius of R=6.378×106 (m). Substitution of the chain rule expression dv/dt=v(dv/dr) as in Example 4 gives

vdvdr=GMr2.

Then integration of both sides with respect to r yields

12v2=GMr+C.

Now v=v0 and r=R when t=0, so C=12v02GM/R, and hence solution for v2 gives

(23)v2=v02+2GM(1r1R).

This implicit solution of Eq. (22) determines the projectile’s velocity v as a function of its distance r from the earth’s center. In particular,

v2>v022GMR,

so v will remain positive provided that v022GM/R. Therefore, the escape velocity from the earth is given by

(24)v0=2GMR.

In Problem 27 we ask you to show that, if the projectile’s initial velocity exceeds 2GM/R, then r(t) as t, so it does, indeed, “escape” from the earth. With the given values of G and the earth’s mass M and radius R, this gives v011,180 (m/s) (about 36,680 ft/s, about 6.95 mi/s, about 25,000 mi/h).

Remark

Equation (24) gives the escape velocity for any other (spherical) planetary body when we use its mass and radius. For instance, when we use the mass M and radius R for the moon given in Example 4, we find that escape velocity from the lunar surface is v02375 m/s. This is just over one-fifth of the escape velocity from the earth’s surface, a fact that greatly facilitates the return trip (“From the Moon to the Earth”).

2.3 Problems

  1. The acceleration of a Maserati is proportional to the difference between 250 km/h and the velocity of this sports car. If this machine can accelerate from rest to 100 km/h in 10 s, how long will it take for the car to accelerate from rest to 200 km/h?

Problems 2 through 8 explore the effects of resistance proportional to a power of the velocity.

  1. Suppose that a body moves through a resisting medium with resistance proportional to its velocity v, so that dv/dt=kv. (a) Show that its velocity and position at time t are given by

    v(t)=v0ekt

    and

    x(t)=x0+(v0k)(1ekt).

    (b) Conclude that the body travels only a finite distance, and find that distance.

  2. Suppose that a motorboat is moving at 40 ft/s when its motor suddenly quits, and that 10 s later the boat has slowed to 20 ft/s. Assume, as in Problem 2, that the resistance it encounters while coasting is proportional to its velocity. How far will the boat coast in all?

  3. Consider a body that moves horizontally through a medium whose resistance is proportional to the square of the velocity v, so that dv/dt=kv2. Show that

    v(t)=v01+v0kt

    and that

    x(t)=x0+1kln (1+v0kt).

    Note that, in contrast with the result of Problem 2, x(t)+ as t+. Which offers less resistance when the body is moving fairly slowly—the medium in this problem or the one in Problem 2? Does your answer seem consistent with the observed behaviors of x(t) as t?

  4. Assuming resistance proportional to the square of the velocity (as in Problem 4), how far does the motorboat of Problem 3 coast in the first minute after its motor quits?

  5. Assume that a body moving with velocity v encounters resistance of the form dv/dt=kv3/2. Show that

    v(t)=4v0(ktv0+2)2

    and that

    x(t)=x0+2kv0(12ktv0+2).

    Conclude that under a 32-power resistance a body coasts only a finite distance before coming to a stop.

  6. Suppose that a car starts from rest, its engine providing an acceleration of 10 ft/s2, while air resistance provides 0.1 ft/s2 of deceleration for each foot per second of the car’s velocity. (a) Find the car’s maximum possible (limiting) velocity. (b) Find how long it takes the car to attain 90% of its limiting velocity, and how far it travels while doing so.

  7. Rework both parts of Problem 7, with the sole difference that the deceleration due to air resistance now is (0.001)v2 ft/s2 when the car’s velocity is v feet per second.

Problems 9 through 12 illustrate resistance proportional to the velocity.

  1. A motorboat weighs 32,000 lb and its motor provides a thrust of 5000 lb. Assume that the water resistance is 100 pounds for each foot per second of the speed v of the boat. Then

    1000dvdt=5000100v.

    If the boat starts from rest, what is the maximum velocity that it can attain?

  2. Falling parachutist A woman bails out of an airplane at an altitude of 10,000 ft, falls freely for 20 s, then opens her parachute. How long will it take her to reach the ground? Assume linear air resistance ρv ft/s2, taking ρ=0.15 without the parachute and ρ=1.5 with the parachute. (Suggestion: First determine her height above the ground and velocity when the parachute opens.)

  3. Falling paratrooper According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but provided some resistance by flapping unopened in the wind. Allegedly he hit the ground at 100 mi/h after falling for 8 s. Test the accuracy of this account. (Suggestion: Find ρ in Eq. (4) by assuming a terminal velocity of 100 mi/h. Then calculate the time required to fall 1200 ft.)

  4. Nuclear waste disposal It is proposed to dispose of nuclear wastes—in drums with weight W=640 lb and volume 8 ft3—by dropping them into the ocean (v0=0). The force equation for a drum falling through water is

    mdvdt=W+B+FR,

    where the buoyant force B is equal to the weight (at 62.5 lb/ft3) of the volume of water displaced by the drum (Archimedes’ principle) and FR is the force of water resistance, found empirically to be 1 lb for each foot per second of the velocity of a drum. If the drums are likely to burst upon an impact of more than 75 ft/s, what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting?

  5. Separate variables in Eq. (12) and substitute u=vρ/g to obtain the upward-motion velocity function given in Eq. (13) with initial condition v(0)=v0.

  6. Integrate the velocity function in Eq. (13) to obtain the upward-motion position function given in Eq. (14) with initial condition y(0)=y0.

  7. Separate variables in Eq. (15) and substitute u=vρ/g to obtain the downward-motion velocity function given in Eq. (16) with initial condition v(0)=v0.

  8. Integrate the velocity function in Eq. (16) to obtain the downward-motion position function given in Eq. (17) with initial condition y(0)=y0.

Problems 17 and 18 apply Eqs. (12)(17) to the motion of a crossbow bolt.

  1. Consider the crossbow bolt of Example 3, shot straight upward from the ground (y=0) at time t=0 with initial velocity v0=49 m/s. Take g=9.8 m/s2 and ρ=0.0011 in Eq. (12). Then use Eqs. (13) and (14) to show that the bolt reaches its maximum height of about 108.47 m in about 4.61 s.

  2. Continuing Problem 17, suppose that the bolt is now dropped (v0=0) from a height of y0=108.47 m. Then use Eqs. (16) and (17) to show that it hits the ground about 4.80 s later with an impact speed of about 43.49 m/s.

Problems 19 through 23 illustrate resistance proportional to the square of the velocity.

  1. A motorboat starts from rest (initial velocity v(0)=v0=0). Its motor provides a constant acceleration of 4 ft/s2, but water resistance causes a deceleration of v2/400 ft/s2. Find v when t=10 s, and also find the limiting velocity as t+ (that is, the maximum possible speed of the boat).

  2. An arrow is shot straight upward from the ground with an initial velocity of 160 ft/s. It experiences both the deceleration of gravity and deceleration v2/800 due to air resistance. How high in the air does it go?

  3. If a ball is projected upward from the ground with initial velocity v0 and resistance proportional to v2, deduce from Eq. (14) that the maximum height it attains is

    ymax=12ρln (1+ρv02g).
  4. Suppose that ρ=0.075 (in fps units, with g=32 ft/s2) in Eq. (15) for a paratrooper falling with parachute open. If he jumps from an altitude of 10,000 ft and opens his parachute immediately, what will be his terminal speed? How long will it take him to reach the ground?

  5. Suppose that the paratrooper of Problem 22 falls freely for 30 s with ρ=0.00075 before opening his parachute. How long will it now take him to reach the ground?

Problems 24 through 30 explore gravitational acceleration and escape velocity.

  1. The mass of the sun is 329,320 times that of the earth and its radius is 109 times the radius of the earth. (a) To what radius (in meters) would the earth have to be compressed in order for it to become a black hole—the escape velocity from its surface equal to the velocity c=3×108 m/s of light? (b) Repeat part (a) with the sun in place of the earth.

  2. (a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity v0 less than escape velocity 2GM/R, then the maximum distance from the center of the earth attained by the projectile is

    rmax=2GMR2GMRv02,

    where M and R are the mass and radius of the earth, respectively. (b) With what initial velocity v0 must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with 90% of escape velocity.

  3. Suppose that you are stranded—your rocket engine has failed—on an asteroid of diameter 3 miles, with density equal to that of the earth with radius 3960 miles. If you have enough spring in your legs to jump 4 feet straight up on earth while wearing your space suit, can you blast off from this asteroid using leg power alone?

    1. Suppose a projectile is launched vertically from the surface r=R of the earth with initial velocity v0=2GM/R, so v02=k2/R where k2=2GM. Solve the differential equation dr/dt=k/r (from Eq. (23) in this section) explicitly to deduce that r(t) as t.

    2. If the projectile is launched vertically with initial velocity v0>2GM/R, deduce that

      drdt=k2r+α>kr.

      Why does it again follow that r(t) as t?

  4. (a) Suppose that a body is dropped (v0=0) from a distance r0>R from the earth’s center, so its acceleration is dv/dt=GM/r2. Ignoring air resistance, show that it reaches the height r<r0 at time

    t=r02GM(rr0r2+r0cos1rr0).

    (Suggestion: Substitute r=r0cos2 θ to evaluate r/(r0r)dr.) (b) If a body is dropped from a height of 1000 km above the earth’s surface and air resistance is neglected, how long does it take to fall and with what speed will it strike the earth’s surface?

  5. Suppose that a projectile is fired straight upward from the surface of the earth with initial velocity v0<2GM/R. Then its height y(t) above the surface satisfies the initial value problem

    d2ydt2=GM(y+R)2;y(0)=0,y(0)=v0.

    Substitute dv/dt=v(dv/dy) and then integrate to obtain

    v2=v022GMyR(R+y)

    for the velocity v of the projectile at height y. What maximum altitude does it reach if its initial velocity is 1 km/s?

  6. In Jules Verne’s original problem, the projectile launched from the surface of the earth is attracted by both the earth and the moon, so its distance r(t) from the center of the earth satisfies the initial value problem

    d2rdt2=GMer2+GMm(Sr)2;r(0)=R,r(0)=v0

    where Me and Mm denote the masses of the earth and the moon, respectively; R is the radius of the earth and S=384,400 km is the distance between the centers of the earth and the moon. To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. Thereafter it is “under the control” of the moon, and falls from there to the lunar surface. Find the minimal launch velocity v0 that suffices for the projectile to make it “From the Earth to the Moon.”

2.3 Application Rocket Propulsion

Suppose that the rocket of Fig. 2.3.5 blasts off straight upward from the surface of the earth at time t=0. We want to calculate its height y and velocity v=dy/dt at time t. The rocket is propelled by exhaust gases that exit (rearward) with constant speed c (relative to the rocket). Because of the combustion of its fuel, the mass m=m(t) of the rocket is variable.

To derive the equation of motion of the rocket, we use Newton’s second law in the form

(1)dPdt=F,

where P is momentum (the product of mass and velocity) and F denotes net external force (gravity, air resistance, etc.). If the mass m of the rocket is constant so m(t)0—when its rockets are turned off or burned out, for instance—then Eq. (1) gives

F=d(mv)dt=mdvdt+dmdtv=mdvdt,

which (with dv/dt=a) is the more familiar form F=ma of Newton’s second law.

But here m is not constant. Suppose m changes to m+Δm and v to v+Δv during the short time interval from t to t+Δt. Then the change in the momentum of the rocket itself is

ΔP(m+Δm)(v+Δv)mv=mΔv+vΔm+ΔmΔv.

FIGURE 2.3.5.

An ascending rocket.

But the system also includes the exhaust gases expelled during this time interval, with mass Δm and approximate velocity vc. Hence the total change in momentum during the time interval Δt is

ΔP(mΔv+vΔm+ΔmΔv)+(Δm)(vc)=mΔv+cΔm+ΔmΔv.

Now we divide by Δt and take the limit as Δt0, so Δm0, assuming continuity of m(t). The substitution of the resulting expression for dP/dt in (1) yields the rocket propulsion equation

(2)mdvdt+cdmdt=F.

If F=FG+FR, where FG=mg is a constant force of gravity and FR=kv is a force of air resistance proportional to velocity, then Eq. (2) finally gives

(3)mdvdt+cdmdt=mgkv.

Constant Thrust

Now suppose that the rocket fuel is consumed at the constant “burn rate” β during the time interval [0,t1], during which time the mass of the rocket decreases from m0 to m1. Thus

(4)m(0)=m0,m(t1)=m1,m(t)=m0βt,dmdt=βfor tt1,

with burnout occurring at time t=t1.

Problem 1 Substitute the expressions in (4) into Eq. (3) to obtain the differential equation

(5)(m0βt)dvdt+kv=βc(m0βt)g.

Solve this linear equation for

(6)v(t)=v0Mk/βgβtβk+(βck+gm0βk)(1Mk/β),

where v0=v(0) and

M=m(t)m0=m0βtm0

denotes the rocket’s fractional mass at time t.

No Resistance

Problem 2 For the case of no air resistance, set k=0 in Eq. (5) and integrate to obtain

(7)v(t)=v0gt+clnm0m0βt.

Because m0βt1=m1, it follows that the velocity of the rocket at burnout (t=t1) is

(8)v1=v(t1)=v0gt1+clnm0m1.

Problem 3 Start with Eq. (7) and integrate to obtain

(9)y(t)=(v0+c)t12gt2cβ(m0βt)lnm0m0βt.

It follows that the rocket’s altitude at burnout is

(10)y1=y(t1)=(v0+c)t112gt12cm1βlnm0m1.

Problem 4 The V-2 rocket that was used to attack London in World War II had an initial mass of 12,850 kg, of which 68.5% was fuel. This fuel burned uniformly for 70 seconds with an exhaust velocity of 2 km/s. Assume it encounters air resistance of 1.45 N per m/s of velocity. Then find the velocity and altitude of the V-2 at burnout under the assumption that it is launched vertically upward from rest on the ground.

Problem 5 Actually, our basic differential equation in (3) applies without qualification only when the rocket is already in motion. However, when a rocket is sitting on its launch pad stand and its engines are turned on initially, it is observed that a certain time interval passes before the rocket actually “blasts off” and begins to ascend. The reason is that if v=0 in (3), then the resulting initial acceleration

dvdt=cmdmdtg

of the rocket may be negative. But the rocket does not descend into the ground; it just “sits there” while (because m is decreasing) this calculated acceleration increases until it reaches 0 and (thereafter) positive values so the rocket can begin to ascend. With the notation introduced to describe the constant-thrust case, show that the rocket initially just “sits there” if the exhaust velocity c is less than m0g/β, and that the time tB which then elapses before actual blastoff is given by

tB=m0gβcβg.

Free Space

Suppose finally that the rocket is accelerating in free space, where there is neither gravity nor resistance, so g=k=0. With g=0 in Eq. (8) we see that, as the mass of the rocket decreases from m0 to m1, its increase in velocity is

(11)Δv=v1v0=clnm0m1.

Note that Δv depends only on the exhaust gas speed c and the initial-to-final mass ratio m0/m1, but does not depend on the burn rate β. For example, if the rocket blasts off from rest (v0=0) and c=5 km/s and m0/m1=20, then its velocity at burnout is v1=5ln 2015 km/s. Thus if a rocket initially consists predominantly of fuel, then it can attain velocities significantly greater than the (relative) velocity of its exhaust gases.

2.4 Numerical Approximation: Euler’s Method

It is the exception rather than the rule when a differential equation of the general form

dydx=f(x, y)

can be solved exactly and explicitly by elementary methods like those discussed in Chapter 1. For example, consider the simple equation

(1)dydx=ex2.

A solution of Eq. (1) is simply an antiderivative of ex2. But it is known that every antiderivative of f(x)=ex2 is a nonelementary function—one that cannot be expressed as a finite combination of the familiar functions of elementary calculus. Hence no particular solution of Eq. (1) is finitely expressible in terms of elementary functions. Any attempt to use the symbolic techniques of Chapter 1 to find a simple explicit formula for a solution of (1) is therefore doomed to failure.

As a possible alternative, an old-fashioned computer plotter—one that uses an ink pen to draw curves mechanically—can be programmed to draw a solution curve that starts at the initial point (x0,y0) and attempts to thread its way through the slope field of a given differential equation y=f(x,y). The procedure the plotter carries out can be described as follows.

Figure 2.4.1 illustrates the result of continuing in this fashion—by a sequence of discrete straight-line steps from one starting point to the next. In this figure we see a polygonal curve consisting of line segments that connect the successive points (x0,y0),(x1,y1),(x2,y2),(x3,y3), However, suppose that each “tiny distance” the pen travels along a slope segment—before the midcourse correction that sends it along a fresh new slope segment—is so very small that the naked eye cannot distinguish the individual line segments constituting the polygonal curve. Then the resulting polygonal curve looks like a smooth, continuously turning solution curve of the differential equation. Indeed, this is (in essence) how most of the solution curves shown in the figures of Chapter 1 were computer generated.

FIGURE 2.4.1.

The first few steps in approximating a solution curve.

Leonhard Euler—the great 18th-century mathematician for whom so many mathematical concepts, formulas, methods, and results are named—did not have a computer plotter, and his idea was to do all this numerically rather than graphically. In order to approximate the solution of the initial value problem

(2)dydx=f(x, y),y(x0)=y0,

we first choose a fixed (horizontal) step size h to use in making each step from one point to the next. Suppose we’ve started at the initial point (x0,y0) and after n steps have reached the point (xn,yn). Then the step from (xn,yn) to the next point (xn+1,yn+1) is illustrated in Fig. 2.4.2. The slope of the direction segment through (xn,yn) is m=f(xn,yn). Hence a horizontal change of h from xn to xn+1 corresponds to a vertical change of m·h=h·f(xn,yn) from yn to yn+1. Therefore the coordinates of the new point (xn+1,yn+1) are given in terms of the old coordinates by

FIGURE 2.4.2.

The step from (xn, yn) to (xn+1, yn+1).

xn+1=xn+h,yn+1=yn+hf(xn, yn).

Given the initial value problem in (2), Euler’s method with step size h consists of starting with the initial point (x0,y0) and applying the formulas

x1=x0+hy1=y0+hf(x0, y0)x2=x1+hy2=y1+hf(x1, y1)x3=x2+hy3=y2+hf(x2, y2)

to calculate successive points (x1,y1), (x2,y2), (x3,y3),  on an approximate solution curve.

However, we ordinarily do not sketch the corresponding polygonal approximation. Instead, the numerical result of applying Euler’s method is the sequence of approximations

y1, y2, y3, , yn,

to the true values

y(x1), y(x2), y(x3), , y(xn),

at the points x1,x2,x3,,xn, of the exact (though unknown) solution y(x) of the initial value problem. These results typically are presented in the form of a table of approximate values of the desired solution.

The iterative formula in (3) tells us how to make the typical step from yn to yn+1 and is the heart of Euler’s method. Although the most important applications of Euler’s method are to nonlinear equations, we first illustrate the method with a simple initial value problem whose exact solution is available, just for the purpose of comparison of approximate and actual solutions.

Example 1

Apply Euler’s method to approximate the solution of the initial value problem

(4)dydx=x+15y,y(0)=3,
  1. first with step size h=1 on the interval [0, 5],

  2. then with step size h=0.2 on the interval [0, 1].

Solution

  1. With x0=0, y0=3, f(x,y)=x+15y, and h=1 the iterative formula in (3) yields the approximate values

    y1=y0+h[x0+15y0]=(3)+(1)[0+15(3)]=3.6,y2=y1+h[x1+15y1]=(3.6)+(1)[1+15(3.6)]=3.32,y3=y2+h[x2+15y2]=(3.32)+(1)[2+15(3.32)]=1.984,y4=y3+h[x3+15y3]=(1.984)+(1)[3+15(1.984)]=0.6192,andy5=y4+h[x4+15y4]=(0.6912)+(1)[4+15(0.6912)]4.7430

    at the points x1=1, x2=2, x3=3, x4=4, and x5=5. Note how the result of each calculation feeds into the next one. The resulting table of approximate values is

    x 0 1 2 3 4 5
    Approx. y 3 3.6 3.32 1.984 0.6912 4.7430

    Figure 2.4.3 shows the graph of this approximation, together with the graphs of the Euler approximations obtained with step sizes h=0.2 and 0.05, as well as the graph of the exact solution

    y(x)=22ex/55x25

    that is readily found using the linear-equation technique of Section 1.5. We see that decreasing the step size increases the accuracy, but with any single approximation, the accuracy decreases with distance from the initial point.

    FIGURE 2.4.3.

    Graphs of Euler approximations with step sizes h=1, h=0.2, and h=0.05.

  2. Starting afresh with x0=0, y0=3, f(x,y)=x+15y, and h=0.2, we get the approximate values

    y1=y0+h[x0+15y0]=(3)+(0.2)[0+15(3)]=3.12,y2=y1+h[x1+15y1]=(3.12)+(0.2)[0.2+15(3.12)]3.205,y3=y2+h[x2+15y2](3.205)+(0.2)[0.4+15(3.205)]3.253,y4=y3+h[x3+15y3](3.253)+(0.2)[0.6+15(3.253)]3.263,y5=y4+h[x4+15y4](3.263)+(0.2)[0.8+15(3.263)]3.234

    at the points x1=0.2, x2=0.4, x3=0.6, x4=0.8, and x5=1. The resulting table of approximate values is

    x 0 0.2 0.4 0.6 0.8 1
    Approx. y 3 3.12 3.205 3.253 3.263 3.234

High accuracy with Euler’s method usually requires a very small step size and hence a larger number of steps than can reasonably be carried out by hand. The application material for this section contains calculator and computer programs for automating Euler’s method. One of these programs was used to calculate the table entries shown in Fig. 2.4.4. We see that 500 Euler steps (with step size h=0.002) from x=0 to x=1 yield values that are accurate to within 0.001.

FIGURE 2.4.4.

Euler approximations with step sizes h=0.2,h=0.02, and h=0.002.

x Approx y with h=0.2 Approx y with h=0.02 Approx y with h=0.002 Actual value of y
0 3.000 3.000 3.000 3.000
0.2 3.120 3.104 3.102 3.102
0.4 3.205 3.172 3.168 3.168
0.6 3.253 3.201 3.196 3.195
0.8 3.263 3.191 3.184 3.183
1 3.234 3.140 3.130 3.129

Example 2

Falling baseball Suppose the baseball of Example 3 in Section 1.3 is simply dropped (instead of being thrown downward) from the helicopter. Then its velocity v(t) after t seconds satisfies the initial value problem

(5)dvdt=320.16v,v(0)=0.

We use Euler’s method with h=1 to track the ball’s increasing velocity at 1-second intervals for the first 10 seconds of fall. With t0=0, v0=0, F(t,v)=320.16v, and h=1 the iterative formula in (3) yields the approximate values

v1=v0+h[320.16v0]=(0)+(1)[320.16(0)]=32,v2=v1+h[320.16v1]=(32)+(1)[320.16(32)]=58.88,v3=v2+h[320.16v2]=(58.88)+(1)[320.16(58.88)]81.46,v4=v3+h[320.16v3]=(81.46)+(1)[320.16(81.46)]100.43,andv5=v4+h[320.16v4]=(100.43)+(1)[320.16(100.43)]116.36.

Continuing in this fashion, we complete the h=1 column of v-values shown in the table of Fig. 2.4.5—where we have rounded off velocity entries to the nearest foot per second. The values corresponding to h=0.1 were calculated using a computer, and we see that they are accurate to within about 1 ft/s. Note also that after 10 seconds the falling ball has attained about 80% of its limiting velocity of 200 ft/s.

Local and Cumulative Errors

There are several sources of error in Euler’s method that may make the approximation yn to y(xn) unreliable for large values of n, those for which xn is not sufficiently close to x0. The error in the linear approximation formula

(6)y(xn+1)yn+hf(xn,yn)=yn+1

FIGURE 2.4.5.

Euler approximations in Example 2 with step sizes h=1 and h=0.1.

t Approx v with h=1 Approx v with h=0.1 Actual value of v
1 32 30 30
2 59 55 55
3 81 77 76
4 100 95 95
5 116 111 110
6 130 124 123
7 141 135 135
8 150 145 144
9 158 153 153
10 165 160 160

is the amount by which the tangent line at (xn,yn) departs from the solution curve through (xn,yn), as illustrated in Fig. 2.4.6. This error, introduced at each step in the process, is called the local error in Euler’s method.

FIGURE 2.4.6.

The local error in Euler’s method.

The local error indicated in Fig. 2.4.6 would be the total error in yn+1 if the starting point yn in (6) were an exact value, rather than merely an approximation to the actual value y(xn). But yn itself suffers from the accumulated effects of all the local errors introduced at the previous steps. Thus the tangent line in Fig. 2.4.6 is tangent to the “wrong” solution curve—the one through (xn,yn) rather than the actual solution curve through the initial point (x0,y0). Figure 2.4.7 illustrates this cumulative error in Euler’s method; it is the amount by which the polygonal stepwise path from (x0,y0) departs from the actual solution curve through (x0,y0).

FIGURE 2.4.7.

The cumulative error in Euler’s method.

FIGURE 2.4.8.

Approximating the solution of dy/dx=x+y, y(0)=1 with successively smaller step sizes.

x y with h=0.1 y with h=0.02 y with h=0.005 y with h=0.001 Actual y
0.1 1.1000 1.1082 1.1098 1.1102 1.1103
0.2 1.2200 1.2380 1.2416 1.2426 1.2428
0.3 1.3620 1.3917 1.3977 1.3993 1.3997
0.4 1.5282 1.5719 1.5807 1.5831 1.5836
0.5 1.7210 1.7812 1.7933 1.7966 1.7974
0.6 1.9431 2.0227 2.0388 2.0431 2.0442
0.7 2.1974 2.2998 2.3205 2.3261 2.3275
0.8 2.4872 2.6161 2.6422 2.6493 2.6511
0.9 2.8159 2.9757 3.0082 3.0170 3.0192
1.0 3.1875 3.3832 3.4230 3.4338 3.4366

The usual way of attempting to reduce the cumulative error in Euler’s method is to decrease the step size h. The table in Fig. 2.4.8 shows the results obtained in approximating the exact solution y(x)=2exx1 of the initial value problem

dydx=x+y,y(0)=1,

using the successively smaller step sizes h=0.1, h=0.02, h=0.005, and h=0.001. We show computed values only at intervals of Δx=0.1. For instance, with h=0.001, the computation required 1000 Euler steps, but the value yn is shown only when n is a multiple of 100, so that xn is an integral multiple of 0.1.

By scanning the columns in Fig. 2.4.8 we observe that, for each fixed step size h, the error yactualyapprox increases as x gets farther from the starting point x0=0. But by scanning the rows of the table we see that for each fixed x, the error decreases as the step size h is reduced. The percentage errors at the final point x=1 range from 7.25% with h=0.1 down to only 0.08% with h=0.001. Thus the smaller the step size, the more slowly does the error grow with increasing distance from the starting point.

The column of data for h=0.1 in Fig. 2.4.8 requires only 10 steps, so Euler’s method can be carried out with a hand-held calculator. But 50 steps are required to reach x=1 with h=0.02, 200 steps with h=0.005, and 1000 steps with h=0.001. A computer is almost always used to implement Euler’s method when more than 10 or 20 steps are required. Once an appropriate computer program has been written, one step size is—in principle—just as convenient as another; after all, the computer hardly cares how many steps it is asked to carry out.

Why, then, do we not simply choose an exceedingly small step size (such as h=1012), with the expectation that very great accuracy will result? There are two reasons for not doing so. The first is obvious: the time required for the computation. For example, the data in Fig. 2.4.8 were obtained using a hand-held calculator that carried out nine Euler steps per second. Thus it required slightly over one second to approximate y(1) with h=0.1 and about 1 min 50 s with h=0.001. But with h=1012 it would require over 3000 years!

The second reason is more subtle. In addition to the local and cumulative errors discussed previously, the computer itself will contribute roundoff error at each stage because only finitely many significant digits can be used in each calculation. An Euler’s method computation with h=0.0001 will introduce roundoff errors 1000 times as often as one with h=0.1. Hence with certain differential equations, h=0.1 might actually produce more accurate results than those obtained with h=0.0001, because the cumulative effect of roundoff error in the latter case might exceed combined cumulative and roundoff error in the case h=0.1.

The “best” choice of h is difficult to determine in practice as well as in theory. It depends on the nature of the function f(x, y) in the initial value problem in (2), on the exact code in which the program is written, and on the specific computer used. With a step size that is too large, the approximations inherent in Euler’s method may not be sufficiently accurate, whereas if h is too small, then roundoff errors may accumulate to an unacceptable degree or the program may require too much time to be practical. The subject of error propagation in numerical algorithms is treated in numerical analysis courses and textbooks.

The computations in Fig. 2.4.8 illustrate the common strategy of applying a numerical algorithm, such as Euler’s method, several times in succession, beginning with a selected number n of subintervals for the first application, then doubling n for each succeeding application of the method. Visual comparison of successive results often can provide an “intuitive feel” for their accuracy. In the next two examples we present graphically the results of successive applications of Euler’s method.

Example 3

Approximate logistic solution The exact solution of the logistic initial value problem

dydx=13y(8y),y(0)=1

is y(x)=8/(1+7e8x/3). Figure 2.4.9 shows both the exact solution curve and approximate solution curves obtained by applying Euler’s method on the interval 0x5 with n=5, n=10, and n=20 subintervals. Each of these “curves” actually consists of line segments joining successive points (xn,yn) and (xn+1,yn+1). The Euler approximation with 5 subintervals is poor, and the approximation with 10 subintervals also overshoots the limiting value y=8 of the solution before leveling off, but with 20 subintervals we obtain fairly good qualitative agreement with the actual behavior of the solution.

FIGURE 2.4.9.

Approximating a logistic solution using Euler’s method with n=5, n=10, and n=20 subintervals.

FIGURE 2.4.10.

Approximating the exact solution y=esinx using Euler’s method with 50, 100, 200, and 400 subintervals.

Example 4

The exact solution of the initial value problem

dydx=ycosx,y(0)=1

is the periodic function y(x)=esin x. Figure 2.4.10 shows both the exact solution curve and approximate solution curves obtained by applying Euler’s method on the interval 0x6π with n=50, n=100, n=200, and n=400 subintervals. Even with this many subintervals, Euler’s method evidently has considerable difficulty keeping up with the oscillations in the actual solution. Consequently, the more accurate methods discussed in succeeding sections are needed for serious numerical investigations.

A Word of Caution

The data shown in Fig. 2.4.8 indicate that Euler’s method works well in approximating the solution of dy/dx=x+y, y(0)=1 on the interval [0, 1]. That is, for each fixed x it appears that the approximate values approach the actual value of y(x) as the step size h is decreased. For instance, the approximate values in the rows corresponding to x=0.3 and x=0.5 suggest that y(0.3)1.40 and y(0.5)1.80, in accord with the actual values shown in the final column of the table.

Example 5, in contrast, shows that some initial value problems are not so well behaved.

Example 5

Cautionary example Use Euler’s method to approximate the solution of the initial value problem

(7)dydx=x2+y2,y(0)=1

on the interval [0, 1].

Solution

Here f(x,y)=x2+y2, so the iterative formula of Euler’s method is

(8)yn+1=yn+h(xn2+yn2).

With step size h=0.1 we obtain

y1=1+(0.1)[(0)2+(1)2]=1.1,y2=1.1+(0.1)[(0.1)2+(1.1)2]=1.222,y3=1.222+(0.1)[(0.2)2+(1.222)2]1.3753,

and so forth. Rounded to four decimal places, the first ten values obtained in this manner are

y1=1.1000y6=2.1995y2=1.2220y7=2.7193y3=1.3753y8=3.5078y4=1.5735y9=4.8023y5=1.8371y10=7.1895

But instead of naively accepting these results as accurate approximations, we decided to use a computer to repeat the computations with smaller values of h. The table in Fig. 2.4.11 shows the results obtained with step sizes h=0.1, h=0.02, and h=0.005. Observe that now the “stability” of the procedure in Example 1 is missing. Indeed, it seems obvious that something is going wrong near x=1.

Figure 2.4.12 provides a graphical clue to the difficulty. It shows a slope field for dy/dx=x2+y2, together with a solution curve through (0, 1) plotted using one of the more accurate approximation methods of the following two sections. It appears that this solution curve may have a vertical asymptote near x=0.97. Indeed, an exact solution using Bessel functions (see Problem 34 in Section 11.4) can be used to show that y(x)+ as x0.969811 (approximately). Although Euler’s method gives values (albeit spurious ones) at x=1, the actual solution does not exist on the entire interval [0, 1]. Moreover, Euler’s method is unable to “keep up” with the rapid changes in y(x) that occur as x approaches the infinite discontinuity near 0.969811.

The moral of Example 5 is that there are pitfalls in the numerical solution of certain initial value problems. Certainly it’s pointless to attempt to approximate a solution on an interval where it doesn’t even exist (or where it is not unique, in which case there’s no general way to predict which way the numerical approximations will branch at a point of nonuniqueness). One should never accept as accurate the results of applying Euler’s method with a single fixed step size h. A second “run” with smaller step size (h/2, say, or h/5, or h/10) may give seemingly consistent results, thereby suggesting their accuracy, or it may—as in Example 5—reveal the presence of some hidden difficulty in the problem. Many problems simply require the more accurate and powerful methods that are discussed in the final two sections of this chapter.

FIGURE 2.4.11.

Attempting to approximate the solution of dy/dx=x2+y2, y(0)=1.

x y with h=0.1 y with h=0.02 y with h=0.005
0.1 1.1000 1.1088 1.1108
0.2 1.2220 1.2458 1.2512
0.3 1.3753 1.4243 1.4357
0.4 1.5735 1.6658 1.6882
0.5 1.8371 2.0074 2.0512
0.6 2.1995 2.5201 2.6104
0.7 2.7193 3.3612 3.5706
0.8 3.5078 4.9601 5.5763
0.9 4.8023 9.0000 12.2061
1.0 7.1895 30.9167 1502.2090

FIGURE 2.4.12.

Solution of dy/dx=x2+y2, y(0)=1.

2.4 Problems

In Problems 1 through 10, an initial value problem and its exact solution y(x) are given. Apply Euler’s method twice to approximate to this solution on the interval [0,12], first with step size h=0.25, then with step size h=0.1. Compare the three-decimal-place values of the two approximations at x=12 with the value y(12) of the actual solution.

  1. y=y, y(0)=2; y(x)=2ex

  2. y=2y, y(0)=12; y(x)=12e2x

  3. y=y+1, y(0)=1; y(x)=2ex1

  4. y=xy, y(0)=1; y(x)=2ex+x1

  5. y=yx1, y(0)=1; y(x)=2+xex

  6. y=2xy, y(0)=2; y(x)=2ex2

  7. y=3x2y, y(0)=3; y(x)=3ex3

  8. y=ey, y(0)=0; y(x)=ln (x+1)

  9. y=14(1+y2), y(0)=1; y(x)=tan14(x+π)

  10. y=2xy2, y(0)=1; y(x)=11x2

Note: The application following this problem set lists illustrative calculator/computer programs that can be used in the remaining problems.

A programmable calculator or a computer will be useful for Problems 11 through 16. In each problem find the exact solution of the given initial value problem. Then apply Euler’s method twice to approximate (to four decimal places) this solution on the given interval, first with step size h=0.01, then with step size h=0.005. Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for x an integral multiple of 0.2. Throughout, primes denote derivatives with respect to x.

  1. y=y2, y(0)=1;0x1

  2. y=12(y1)2, y(0)=2;0x1

  3. yy=2x3, y(1)=3;1x2

  4. xy=y2, y(1)=1;1x2

  5. xy=3x2y, y(2)=3;2x3

  6. y2y=2x5, y(2)=3;2x3

A computer with a printer is required for Problems 17 through 24. In these initial value problems, use Euler’s method with step sizes h=0.1, 0.02, 0.004, and 0.0008 to approximate to four decimal places the values of the solution at ten equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to x.

  1. y=x2+y2, y(0)=0; 0x1

  2. y=x2y2, y(0)=1; 0x2

  3. y=x+y, y(0)=1; 0x2

  4. y=x+y3, y(0)=1; 0x2

  5. y=ln y, y(1)=2; 1x2

  6. y=x2/3+y2/3, y(0)=1; 0x2

  7. y=sin x+cos y, y(0)=0; 0x1

  8. y=x1+y2, y(1)=1; 1x1

  9. Falling parachutist You bail out of the helicopter of Example 2 and immediately pull the ripcord of your parachute. Now k=1.6 in Eq. (5), so your downward velocity satisfies the initial value problem

    dvdt=321.6v,v(0)=0

    (with t in seconds and v in ft/sec). Use Euler’s method with a programmable calculator or computer to approximate the solution for 0t2, first with step size h=0.01 and then with h=0.005, rounding off approximate v-values to one decimal place. What percentage of the limiting velocity 20 ft/sec has been attained after 1 second? After 2 seconds?

  10. Deer population Suppose the deer population P(t) in a small forest initially numbers 25 and satisfies the logistic equation

    dPdt=0.0225P0.0003P2

    (with t in months). Use Euler’s method with a programmable calculator or computer to approximate the solution for 10 years, first with step size h=1 and then with h=0.5, rounding off approximate P-values to integral numbers of deer. What percentage of the limiting population of 75 deer has been attained after 5 years? After 10 years?

Use Euler’s method with a computer system to find the desired solution values in Problems 27 and 28. Start with step size h=0.1, and then use successively smaller step sizes until successive approximate solution values at x=2 agree rounded off to two decimal places.

  1. y=x2+y21, y(0)=0; y(2)=?

  2. y=x+12y2, y(2)=0; y(2)=?

Problems 29 through 31 illustrate the unreliability of Euler’s method near a discontinuity of the solution.

  1. Consider the initial value problem

    7xdydx+y=0,y(1)=1.

    (a) Solve this problem for the exact solution

    y(x)=1x1/7,

    which has an infinite discontinuity at x=0. (b) Apply Euler’s method with step size h=0.15 to approximate this solution on the interval 1x0.5. Note that, from these data alone, you might not suspect any difficulty near x=0. The reason is that the numerical approximation “jumps across the discontinuity” to another solution of 7xy+y=0 for x>0. (c) Finally, apply Euler’s method with step sizes h=0.03 and h=0.006, but still printing results only at the original points x=1.00, 0.85, 0.70, , 1.20, 1.35. and 1.50. Would you now suspect a discontinuity in the exact solution?

  2. Apply Euler’s method with successively smaller step sizes on the interval [0, 2] to verify empirically that the solution of the initial value problem

    dydx=x2+y2,y(0)=0

    has a vertical asymptote near x=2.003147. (Contrast this with Example 2, in which y(0)=1.)

  3. The general solution of the equation

    dydx=(1+y2)cosx

    is y(x)=tan(C+sin x). With the initial condition y(0)=0 the solution y(x)=tan(sin x) is well behaved. But with y(0)=1 the solution y(x)=tan(14π+sin x) has a vertical asymptote at x=sin1(π/4)0.90334. Use Euler’s method to verify this fact empirically.

2.4 Application Implementing Euler’s Method

Construction of a calculator or computer program to implement a numerical algorithm can sharpen one’s understanding of the algorithm. Figure 2.4.13 lists TI-84 Plus and Python programs implementing Euler’s method to approximate the solution of the initial value problem

dydx=x+y,y(0)=1

considered in this section. The comments provided in the final column should make these programs intelligible even if you have little familiarity with the Python and TI calculator programming languages. Python—freely available on the Internet—is a general-purpose computing language widely used in education and industry. In particular, its many add-on packages make Python a powerful scientific computing platform, with numerical, graphic, and symbolic capabilities approaching those of commercial systems such as Matlab, Mathematica, and Maple. Further, Python code was designed with an emphasis on readability, making it well-suited to expressing basic mathematical algorithms.

FIGURE 2.4.13.

TI-84 Plus and Python Euler’s method programs.

TI-84 Plus Python Comment
PROGRAM: EULER # Program EULER Program title
:10→N N = 10 Number of steps
:0→X X = 0.0 Initial x
:1→Y Y = 1.0 Initial y
:1→T X1 = 1.0 Final x
:(T-X)/N→H H = (X1-X)/N Step size
:For(I,1,N) for I in range(N): Begin loop
:X+Y→F F = X + Y Function value
:Y+H*F→Y Y = Y + H*F Euler iteration
:X+H→X X = X + H New x
:Disp X,Y print (X,Y) Display results
:End # END End loop

To increase the number of steps (and thereby decrease the step size) you need only change the value of N specified in the first line of the program. To apply Euler’s method to a different equation dy/dx=f(x,y), you need change only the single line that calculates the function value F.

Any other procedural programming language (such as FORTRAN or C++) would follow the pattern illustrated by the parallel lines of TI-84 Plus and Python code in Fig. 2.4.13. Some of the modern functional programming languages mirror standard mathematical notation even more closely. Figure 2.4.14 shows a Matlab implementation of Euler’s method. The euler function takes as input the initial value x, the initial value y, the final value x1 of x, and the desired number of subintervals.

function yp = f(x,y)
yp = x + y;                % yp = y’
function [X,Y] = euler(x,y,x1,n)
h = (x1 - x)/n;            % step size
X = x;                     % initial x
Y = y;                     % initial y
for i = 1:n                % begin loop
    y = y + h*f(x,y);      % Euler iteration
    x = x + h;             % new x
    X = [X; x];            % update x-column
    Y = [Y; y];            % update y-column
    end                    % end loop

FIGURE 2.4.14.

Matlab implementation of Euler’s method.

For instance, the Matlab command

[X, Y] = euler(0, 1, 1, 10)

then generates the xn- and yn-data shown in the first two columns of the table of Fig. 2.4.8.

You should begin this project by implementing Euler’s method with your own calculator or computer system. Test your program by first applying it to the initial value problem in Example 1, then to some of the problems for this section.

Famous Numbers Investigation

The following problems describe the numbers e2.71828, ln 20.69315, and π3.14159 as specific values of solutions of certain initial value problems. In each case, apply Euler’s method with n=50, 100, 200,  subintervals (doubling n each time). How many subintervals are needed to obtain—twice in succession—the correct value of the target number rounded to three decimal places?

  1. The number e=y(1), where y(x) is the solution of the initial value problem dy/dx=y, y(0)=1.

  2. The number ln 2=y(2), where y(x) is the solution of the initial value problem dy/dx=1/x, y(1)=0.

  3. The number π=y(1), where y(x) is the solution of the initial value problem dy/dx=4/(1+x2), y(0)=0.

Also explain in each problem what the point is—why the indicated famous number is the expected numerical result.

2.5 A Closer Look at the Euler Method

The Euler method as presented in Section 2.4 is not often used in practice, mainly because more accurate methods are available. But Euler’s method has the advantage of simplicity, and a careful study of this method yields insights into the workings of more accurate methods, because many of the latter are extensions or refinements of the Euler method.

To compare two different methods of numerical approximation, we need some way to measure the accuracy of each. Theorem 1 tells what degree of accuracy we can expect when we use Euler’s method.

Remark

The error

yactualyapprox=y(xn)yn

in (2) denotes the [cumulative] error in Euler’s method after n steps in the approximation, exclusive of roundoff error (as though we were using a perfect machine that made no roundoff errors). The theorem can be summarized by saying that the error in Euler’s method is of order h; that is, the error is bounded by a [predetermined] constant C multiplied by the step size h. It follows, for instance, that (on a given closed interval) halving the step size cuts the maximum error in half; similarly, with step size h/10 we get 10 times the accuracy (that is, 1/10 the maximum error) as with step size h. Consequently, we can—in principle—get any degree of accuracy we want by choosing h sufficiently small.

We will omit the proof of this theorem, but one can be found in Chapter 7 of G. Birkhoff and G.-C. Rota, Ordinary Differential Equations, 4th ed. (New York: John Wiley, 1989). The constant C deserves some comment. Because C tends to increase as the maximum value of |y(x)| on [a, b] increases, it follows that C must depend in a fairly complicated way on y, and actual computation of a value of C such that the inequality in (2) holds is usually impractical. In practice, the following type of procedure is commonly employed.

  1. Apply Euler’s method to the initial value problem in (1) with a reasonable value of h.

  2. Repeat with h/2, h/4, and so forth, at each stage halving the step size for the next application of Euler’s method.

  3. Continue until the results obtained at one stage agree—to an appropriate number of significant digits—with those obtained at the previous stage. Then the approximate values obtained at this stage are considered likely to be accurate to the indicated number of significant digits.

Example 1

Carry out this procedure with the initial value problem

(3)dydx=2xy1+x2,y(0)=1

to approximate accurately the value y(1) of the solution at x=1.

Solution

Using an Euler method program, perhaps one of those listed in Figs. 2.4.13 and 2.4.14, we begin with a step size h=0.04 requiring n=25 steps to reach x=1. The table in Fig. 2.5.1 shows the approximate values of y(1) obtained with successively smaller values of h. The data suggest that the true value of y(1) is exactly 0.5. Indeed, the exact solution of the initial value problem in (3) is y(x)=1/(1+x2), so the true value of y(1) is exactly 12.

FIGURE 2.5.1.

Table of values in Example 1.

h Approximate y(1) Actual y(1) |Error|/h
0.04 0.50451 0.50000 0.11
0.02 0.50220 0.50000 0.11
0.01 0.50109 0.50000 0.11
0.005 0.50054 0.50000 0.11
0.0025 0.50027 0.50000 0.11
0.00125 0.50013 0.50000 0.10
0.000625 0.50007 0.50000 0.11
0.0003125 0.50003 0.50000 0.10

The final column of the table in Fig. 2.5.1 displays the ratio of the magnitude of the error to h; that is, |yactualyapprox|/h. Observe how the data in this column substantiate Theorem 1—in this computation, the error bound in (2) appears to hold with a value of C slightly larger than 0.1.

An Improvement in Euler’s Method

As Fig. 2.5.2 shows, Euler’s method is rather unsymmetrical. It uses the predicted slope k=f(xn,yn) of the graph of the solution at the left-hand endpoint of the interval [xn,xn+h] as if it were the actual slope of the solution over that entire interval. We now turn our attention to a way in which increased accuracy can easily be obtained; it is known as the improved Euler method.

Given the initial value problem

(4)dydx=f(x, y),y(x0)=y0,

suppose that after carrying out n steps with step size h we have computed the approximation yn to the actual value y(xn) of the solution at xn=x0+nh. We can use the Euler method to obtain a first estimate—which we now call un+1 rather than yn+1—of the value of the solution at xn+1=xn+h. Thus

un+1=yn+hf(xn, yn)=yn+hk1.

FIGURE 2.5.2.

True and predicted values in Euler’s method.

Now that un+1y(xn+1) has been computed, we can take

k2=f(xn+1, un+1)

as a second estimate of the slope of the solution curve y=y(x) at x=xn+1.

Of course, the approximate slope k1=f(xn,yn) at x=xn has already been calculated. Why not average these two slopes to obtain a more accurate estimate of the average slope of the solution curve over the entire subinterval [xn,xn+1]? This idea is the essence of the improved Euler method. Figure 2.5.3 shows the geometry behind this method.

FIGURE 2.5.3.

The improved Euler method: Average the slopes of the tangent lines at (xn, yn) and (xn+1, un+1).

Remark

The final formula in (5) takes the “Euler form”

yn+1=yn+hk

if we write

k=k1+k22

for the approximate average slope on the interval [xn,xn+1].

The improved Euler method is one of a class of numerical techniques known as predictor-corrector methods. First a predictor un+1 of the next y-value is computed; then it is used to correct itself. Thus the improved Euler method with step size h consists of using the predictor

(6)un+1=yn+hf(xn, yn)

and the corrector

(7)yn+1=yn+h12[f(xn, yn)+f(xn+1, un+1)]

iteratively to compute successive approximations y1, y2, y2,  to the values y(x1), y(x2), y(x3),  of the actual solution of the initial value problem in (4).

Remark

Each improved Euler step requires two evaluations of the function f(x, y), as compared with the single function evaluation required for an ordinary Euler step. We naturally wonder whether this doubled computational labor is worth the trouble.

Answer

Under the assumption that the exact solution y=y(x) of the initial value problem in (4) has a continuous third derivative, it can be proved—see Chapter 7 of Birkhoff and Rota—that the error in the improved Euler method is of order h2. This means that on a given bounded interval [a, b], each approximate value yn satisfies the inequality

(8)|y(xn)yn|Ch2,

where the constant C does not depend on h. Because h2 is much smaller than h if h itself is small, this means that the improved Euler method is more accurate than Euler’s method itself. This advantage is offset by the fact that about twice as many computations are required. But the factor h2 in (8) means that halving the step size results in 1/4 the maximum error, and with step size h/10 we get 100 times the accuracy (that is, 1/100 the maximum error) as with step size h.

Example 2

Figure 2.4.8 shows results of applying Euler’s method to the initial value problem

(9)dydx=x+y,y(0)=1

with exact solution y(x)=2exx1. With f(x,y)=x+y in Eqs. (6) and (7), the predictor-corrector formulas for the improved Euler method are

un+1=yn+h(xn+yn),yn+1=yn+h12[(xn+yn)+(xn+1+un+1)].

With step size h=0.1 we calculate

u1=1+(0.1)(0+1)=1.1,y1=1+(0.05)[(0+1)+(0.1+1.1)]=1.11,u2=1.11+(0.1)(0.1+1.11)=1.231,y2=1.11+(0.05)[(0.1+1.11)+(0.2+1.231)]=1.24205,

and so forth. The table in Fig. 2.5.4 compares the results obtained using the improved Euler method with those obtained previously using the “unimproved” Euler method. When the same step size h=0.1 is used, the error in the Euler approximation to y(1) is 7.25%, but the error in the improved Euler approximation is only 0.24%.

FIGURE 2.5.4.

Euler and improved Euler approximations to the solution of dy/dx=x+y, y(0)=1.

x Euler Method, h=0.1 Values of y Euler Method, h=0.005 Values of y Improved Euler, h=0.1 Values of y Actual y
0.1 1.1000 1.1098 1.1100 1.1103
0.2 1.2200 1.2416 1.2421 1.2428
0.3 1.3620 1.3977 1.3985 1.3997
0.4 1.5282 1.5807 1.5818 1.5836
0.5 1.7210 1.7933 1.7949 1.7974
0.6 1.9431 2.0388 2.0409 2.0442
0.7 2.1974 2.3205 2.3231 2.3275
0.8 2.4872 2.6422 2.6456 2.6511
0.9 2.8159 3.0082 3.0124 3.0192
1.0 3.1875 3.4230 3.4282 3.4366

FIGURE 2.5.5.

Improved Euler approximation to the solution of Eq. (9) with step size h=0.005.

x Improved Euler, Approximate y Actual y
0.0 1.00000 1.00000
0.1 1.11034 1.11034
0.2 1.24280 1.24281
0.3 1.39971 1.39972
0.4 1.58364 1.58365
0.5 1.79744 1.79744
0.6 2.04423 2.04424
0.7 2.32749 2.32751
0.8 2.65107 2.65108
0.9 3.01919 3.01921
1.0 3.43654 3.43656

Indeed, the improved Euler method with h=0.1 is more accurate (in this example) than the original Euler method with h=0.005. The latter requires 200 evaluations of the function f(x, y), but the former requires only 20 such evaluations, so in this case the improved Euler method yields greater accuracy with only about one-tenth the work.

Figure 2.5.5 shows the results obtained when the improved Euler method is applied to the initial value problem in (9) using step size h=0.005. Accuracy of five significant figures is apparent in the table. This suggests that, in contrast with the original Euler method, the improved Euler method is sufficiently accurate for certain practical applications—such as plotting solution curves.

FIGURE 2.5.6.

Improved Euler approximation to y(1) for dy/dx=2xy/(1+x2), y(0)=1.

h Improved Euler Approximation to y(1) Error |Error|/h2
0.04 0.500195903 0.000195903 0.12
0.02 0.500049494 0.000049494 0.12
0.01 0.500012437 0.000012437 0.12
0.005 0.500003117 0.000003117 0.12
0.0025 0.500000780 0.000000780 0.12
0.00125 0.500000195 0.000000195 0.12
0.000625 0.500000049 0.000000049 0.12
0.0003125 0.500000012 0.000000012 0.12

An improved Euler program (similar to the ones listed in the project material for this section) was used to compute approximations to the exact value y(1)=0.5 of the solution y(x)=1/(1+x2) of the initial value problem

(3)dydx=2xy1+x2,y(0)=1

of Example 1. The results obtained by successively halving the step size appear in the table in Fig. 2.5.6. Note that the final column of this table impressively corroborates the form of the error bound in (8), and that each halving of the step size reduces the error by a factor of almost exactly 4, as should happen if the error is proportional to h2.

In the following two examples we exhibit graphical results obtained by employing this strategy of successively halving the step size, and thus doubling the number of subintervals of a fixed interval on which we are approximating a solution.

Example 3

In Example 3 of Section 2.4 we applied Euler’s method to the logistic initial value problem

dydx=13y(8y),y(0)=1.

Figure 2.4.9 shows an obvious difference between the exact solution y(x)= 8/(1+7e8x/3) and the Euler approximation on 0x5 using n=20 subintervals. Figure 2.5.7 shows approximate solution curves plotted using the improved Euler’s method.

The approximation with five subintervals is still bad—perhaps worse! It appears to level off considerably short of the actual limiting population M=8. You should carry out at least the first two improved Euler steps manually to see for yourself how it happens that, after increasing appropriately during the first step, the approximate solution decreases in the second step rather than continuing to increase (as it should). In the project for this section we ask you to show empirically that the improved Euler approximate solution with step size h=1 levels off at y4.3542.

In contrast, the approximate solution curve with n=20 subintervals tracks the exact solution curve rather closely, and with n=40 subintervals the exact and approximate solution curves are indistinguishable in Fig. 2.5.7. The table in Fig. 2.5.8 indicates that the improved Euler approximation with n=200 subintervals is accurate rounded to three decimal places (that is, four significant digits) on the interval 0x5. Because discrepancies in the fourth significant digit are not visually apparent at the resolution of an ordinary computer screen, the improved Euler method (using several hundred subintervals) is considered adequate for many graphical purposes.

FIGURE 2.5.7.

Approximating a logistic solution using the improved Euler method with n=5, n=10, n=20, and n=40 subintervals.

FIGURE 2.5.8.

Using the improved Euler method to approximate the actual solution of the initial value problem in Example 3.

x Actual y(x) Improved Euler with n=200
0 1.0000 1.0000
1 5.3822 5.3809
2 7.7385 7.7379
3 7.9813 7.9812
4 7.9987 7.9987
5 7.9999 7.9999

Example 4

In Example 4 of Section 2.4 we applied Euler’s method to the initial value problem

dydx=ycosx,y(0)=1.

Figure 2.4.10 shows obvious visual differences between the periodic exact solution y(x)=esin x and the Euler approximations on 0x6π with as many as n=400 subintervals.

FIGURE 2.5.9.

Approximating the exact solution y=esin x using the improved Euler method with n=50, 100, and 200 subintervals.

Figure 2.5.9 shows the exact solution curve and approximate solution curves plotted using the improved Euler method with n=50, n=100, and n=200 subintervals. The approximation obtained with n=200 is indistinguishable from the exact solution curve, and the approximation with n=100 is only barely distinguishable from it.

Although Figs. 2.5.7 and 2.5.9 indicate that the improved Euler method can provide accuracy that suffices for many graphical purposes, it does not provide the higher-precision numerical accuracy that sometimes is needed for more careful investigations. For instance, consider again the initial value problem

dydx=2xy1+x2,y(0)=1

of Example 1. The final column of the table in Fig. 2.5.6 suggests that, if the improved Euler method is used on the interval 0x1 with n subintervals and step size h=1/n, then the resulting error E in the final approximation yny(1) is given by

E=|y(1)yn|(0.12)h2=0.12n2.

If so, then 12-place accuracy (for instance) in the value y(1) would require that (0.12)n2<5×1013, which means that n489,898. Thus, roughly half a million steps of length h0.000002 would be required. Aside from the possible impracticality of this many steps (using available computational resources), the roundoff error resulting from so many successive steps might well overwhelm the cumulative error predicted by theory (which assumes exact computations in each separate step). Consequently, still more accurate methods than the improved Euler method are needed for such high-precision computations. Such a method is presented in Section 2.6.

2.5 Problems

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval [0, 0.5] with step size h=0.1. Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points x=0.1, 0.2, 0.3, 0.4, 0.5.

  1. y=y, y(0)=2; y(x)=2ex

  2. y=2y, y(0)=12; y(x)=12e2x

  3. y=y+1, y(0)=1; y(x)=2ex1

  4. y=xy, y(0)=1; y(x)=2ex+x1

  5. y=yx1, y(0)=1; y(x)=2+xex

  6. y=2xy, y(0)=2; y(x)=2ex2

  7. y=3x2y, y(0)=3; y(x)=3ex3

  8. y=ey, y(0)=0; y(x)=ln (x+1)

  9. y=14(1+y2), y(0)=1; y(x)=tan14(x+π)

  10. y=2xy2, y(0)=1; y(x)=11x2

Note: The application following this problem set lists illustrative calculator/computer programs that can be used in Problems 11 through 24.

A programmable calculator or a computer will be useful for Problems 11 through 16. In each problem find the exact solution of the given initial value problem. Then apply the improved Euler method twice to approximate (to five decimal places) this solution on the given interval, first with step size h=0.01, then with step size h=0.005. Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximations, for x an integral multiple of 0.2. Throughout, primes denote derivatives with respect to x.

  1. y=y2, y(0)=1; 0x1

  2. y=12(y1)2, y(0)=2; 0x1

  3. yy=2x3, y(1)=3; 1x2

  4. xy=y2, y(1)=1; 1x2

  5. xy=3x2y, y(2)=3; 2x3

  6. y2y=2x5, y(2)=3; 2x3

A computer with a printer is required for Problems 17 through 24. In these initial value problems, use the improved Euler method with step sizes h=0.1, 0.02, 0.004, and 0.0008 to approximate to five decimal places the values of the solution at ten equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to x.

  1. y=x2+y2, y(0)=0; 0x1

  2. y=x2y2, y(0)=1; 0x2

  3. y=x+y, y(0)=1; 0x2

  4. y=x+y3, y(0)=1; 0x2

  5. y=ln y, y(1)=2; 1x2

  6. y=x2/3+y2/3, y(0)=1; 0x2

  7. y=sin x+cos y, y(0)=0; 0x1

  8. y=x1+y2, y(1)=1; 1x1

  9. Falling parachutist As in Problem 25 of Section 2.4, you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem

    dvdt=321.6v,v(0)=0

    (with t in seconds and v in ft/s). Use the improved Euler method with a programmable calculator or computer to approximate the solution for 0t2, first with step size h=0.01 and then with h=0.005, rounding off approximate v-values to three decimal places. What percentage of the limiting velocity 20 ft/s has been attained after 1 second? After 2 seconds?

  10. Deer population As in Problem 26 of Section 2.4, suppose the deer population P(t) in a small forest initially numbers 25 and satisfies the logistic equation

    dPdt=0.0225P0.0003P2

    (with t in months). Use the improved Euler method with a programmable calculator or computer to approximate the solution for 10 years, first with step size h=1 and then with h=0.5, rounding off approximate P-values to three decimal places. What percentage of the limiting population of 75 deer has been attained after 5 years? After 10 years?

Decreasing Step Size

Use the improved Euler method with a computer system to find the desired solution values in Problems 27 and 28. Start with step size h=0.1, and then use successively smaller step sizes until successive approximate solution values at x=2 agree rounded off to four decimal places.

  1. y=x2+y21, y(0)=0; y(2)=?

  2. y=x+12y2, y(2)=0; y(2)=?

  3. Velocity-proportional resistance Consider the crossbow bolt of Example 2 in Section 2.3, shot straight upward from the ground with an initial velocity of 49 m/s. Because of linear air resistance, its velocity function v(t) satisfies the initial value problem

    dυdt=(0.04)υ9.8,υ(0)=49

    with exact solution v(t)=294et/25245. Use a calculator or computer implementation of the improved Euler method to approximate v(t) for 0t10 using both n=50 and n=100 subintervals. Display the results at intervals of 1 second. Do the two approximations—each rounded to two decimal places—agree both with each other and with the exact solution? If the exact solution were unavailable, explain how you could use the improved Euler method to approximate closely (a) the bolt’s time of ascent to its apex (given in Section 2.3 as 4.56 s) and (b) its impact velocity after 9.41 s in the air.

  4. Square-proportional resistance Consider now the crossbow bolt of Example 3 in Section 2.3. It still is shot straight upward from the ground with an initial velocity of 49 m/s, but because of air resistance proportional to the square of its velocity, its velocity function v(t) satisfies the initial value problem

    dvdt=(0.0011)v|v|9.8,v(0)=49.

    The symbolic solution discussed in Section 2.3 required separate investigations of the bolt’s ascent and its descent, with v(t) given by a tangent function during ascent and by a hyperbolic tangent function during descent. But the improved Euler method requires no such distinction. Use a calculator or computer implementation of the improved Euler method to approximate v(t) for 0t10 using both n=100 and n=200 subintervals. Display the results at intervals of 1 second. Do the two approximations—each rounded to two decimal places—agree with each other? If an exact solution were unavailable, explain how you could use the improved Euler method to approximate closely (a) the bolt’s time of ascent to its apex (given in Section 2.3 as 4.61 s) and (b) its impact velocity after 9.41 s in the air.

2.5 Application Improved Euler Implementation

Figure 2.5.10 lists TI-84 Plus and Python programs implementing the improved Euler method to approximate the solution of the initial value problem

dydx=x+y,y(0)=1

considered in Example 2 of this section. The comments provided in the final column should make these programs intelligible even if you have little familiarity with the Python and TI programming languages.

To apply the improved Euler method to a differential equation dy/dx=f(x,y), one need only replace x+y throughout with the desired expression. To increase the number of steps (and thereby decrease the step size) one need only change the value of N specified in the second line of the program.

FIGURE 2.5.10.

TI-84 Plus and Python improved Euler programs.

TI-84 Plus Python Comment
PROGRAM: IMPEULER # Program IMPEULER Program title
  def F(X,Y): return X + Y Define function f
:10→N N = 10 No. of steps
:0→X X = 0.0 Initial x
:1→Y Y = 1.0 Initial y
:1→T X1 = 1.0 Final x
:(T-X)/N→H H = (X1-X)/N Step size
:For(I,1,N) for I in range(N): Begin loop
:Y→Z Y0 = Y Save previous y
:X+Y→K K1 = F(X,Y) First slope
:Z+H*K→Y Y = Y0 + H*K1 Predictor
:X+H→X X = X + H New x
:X+Y→L K2 = F(X,Y) Second slope
:(K+L)/2→K K = (K1 + K2)/2 Average slope
:Z+H*K→Y Y = Y0 + H*K Corrector
:Disp X,Y print (X,Y) Display results
:End # END End loop

Figure 2.5.11 exhibits one Matlab implementation of the improved Euler method. The impeuler function takes as input the initial value x, the initial value y, the final value x1 of x, and the desired number n of subintervals. As output it produces the resulting column vectors x and y of x- and y-values. For instance, the Matlab command

[X, Y] = impeuler(0, 1, 1, 10)

then generates the first and fourth columns of data shown in Fig. 2.5.4.

function yp = f(x,y)
yp = x + y;                   % yp = y’
function [X,Y] = impeuler(x,y,x1,n)
h = (x1 - x)/n;               % step size
X = x;                        % initial x
Y = y;                        % initial y
for i = 1:n;                  % begin loop
    k1 = f(x,y);              % first slope
    k2 = f(x+h,y+h*k1);       % second slope
    k = (k1 + k2)/2;;         % average slope
    x = x + h;                % new x
    y = y + h*k;              % new y
    X = [X;x];                % update x-column
    Y = [Y;y];                % update y-column
    end                       % end loop

FIGURE 2.5.11.

Matlab implementation of improved Euler method.

You should begin this project by implementing the improved Euler method with your own calculator or computer system. Test your program by applying it first to the initial value problem of Example 1, then to some of the problems for this section.

Famous Numbers Revisited

The following problems describe the numbers e2.7182818, ln 20.6931472, and π3.1415927 as specific values of certain initial value problems. In each case, apply the improved Euler method with n=10, 20, 40,  subintervals (doubling n each time). How many subintervals are needed to obtain—twice in succession—the correct value of the target number rounded to five decimal places?

  1. The number e=y(1), where y(x) is the solution of the initial value problem dy/dx=y, y(0)=1.

  2. The number ln 2=y(2), where y(x) is the solution of the initial value problem dy/dx=1/x, y(1)=0.

  3. The number π=y(1), where y(x) is the solution of the initial value problem dy/dx=4/(1+x2), y(0)=0.

Logistic Population Investigation

Apply your improved Euler program to the initial value problem dy/dx=13y(8y), y(0)=1 of Example 3. In particular, verify (as claimed) that the approximate solution with step size h=1 levels off at y4.3542 rather than at the limiting value y=8 of the exact solution. Perhaps a table of values for 0x100 will make this apparent.

For your own logistic population to investigate, consider the initial value problem

dydx=1ny(my),y(0)=1

where m and n are (for instance) the largest and smallest nonzero digits in your student ID number. Does the improved Euler approximation with step size h=1 level off at the “correct” limiting value of the exact solution? If not, find a smaller value of h so that it does.

Periodic Harvesting and Restocking

The differential equation

dydt=ky(My)hsin(2πtP)

models a logistic population that is periodically harvested and restocked with period P and maximal harvesting/restocking rate h. A numerical approximation program was used to plot the typical solution curves for the case k=M=h=P=1 that are shown in Fig. 2.5.12. This figure suggests—although it does not prove—the existence of a threshold initial population such that

FIGURE 2.5.12.

Solution curves of dy/dt=y(1y)sin 2πt.

  • Beginning with an initial population above this threshold, the population oscillates (perhaps with period P?) around the (unharvested) stable limiting population y(t)M, whereas

  • The population dies out if it begins with an initial population below this threshold.

Use an appropriate plotting utility to investigate your own logistic population with periodic harvesting and restocking (selecting typical values of the parameters k, M, h, and P). Do the observations indicated here appear to hold for your population?

2.6 The Runge–Kutta Method

We now discuss a method for approximating the solution y=y(x) of the initial value problem

(1)dydx=f(x, y),y(x0)=y0

that is considerably more accurate than the improved Euler method and is more widely used in practice than any of the numerical methods discussed in Sections 2.4 and 2.5. It is called the Runge–Kutta method, after the German mathematicians who developed it, Carl Runge (1856–1927) and Wilhelm Kutta (1867–1944).

With the usual notation, suppose that we have computed the approximations y1, y2, y3, , yn to the actual values y(x1), y(x2), y(x3), , y(xn) and now want to compute yn+1y(xn+1). Then

(2)y(xn+1)y(xn)=xnxn+1y(x)dx=xnxn+hy(x)dx

by the fundamental theorem of calculus. Next, Simpson’s rule for numerical integration yields

(3)y(xn+1)y(xn)h6[y(xn)+4y(xn+h2)+y(xn+1)].

Hence we want to define yn+1 so that

(4)yn+1yn+h6[y(xn)+2y(xn+h2)+2y(xn+h2)+y(xn+1)];

we have split 4y(xn+12h) into a sum of two terms because we intend to approximate the slope y(xn+12h) at the midpoint xn+12h of the interval [xn,xn+1] in two different ways.

On the right-hand side in (4), we replace the [true] slope values y(xn), y(xn+12h), y(xn+12h), and y(xn+1), respectively, with the following estimates.

(5a)k1=f(xn, yn)
(5b)k2=f(xn+12h, yn+12hk1)
(5c)k3=f(xn+12h, yn+12hk2)
(5d)k4=f(xn+1, yn+hk3)
(6)yn+1=yn+h6(k1+2k2+2k3+k4).

The use of this formula to compute the approximations y1, y2, y3,  successively constitutes the Runge–Kutta method. Note that Eq. (6) takes the “Euler form”

yn+1=yn+hk

if we write

(7)k=16(k1+2k2+2k3+k4)

for the approximate average slope on the interval [xn,xn+1].

The Runge–Kutta method is a fourth-order method—it can be proved that the cumulative error on a bounded interval [a, b] with a=x0 is of order h4. (Thus the iteration in (6) is sometimes called the fourth-order Runge–Kutta method because it is possible to develop Runge–Kutta methods of other orders.) That is,

(8)|y(xn)yn|Ch4,

where the constant C depends on the function f(x, y) and the interval [a, b], but does not depend on the step size h. The following example illustrates this high accuracy in comparison with the lower-order accuracy of our previous numerical methods.

Example 1

We first apply the Runge–Kutta method to the illustrative initial value problem

(9)dydx=x+y,y(0)=1

that we considered in Fig. 2.4.8 of Section 2.4 and again in Example 2 of Section 2.5. The exact solution of this problem is y(x)=2exx1. To make a point we use h=0.5, a larger step size than in any previous example, so only two steps are required to go from x=0 to x=1.

In the first step we use the formulas in (5) and (6) to calculate

k1=0+1=1,k2=(0+0.25)+(1+(0.25)(1))=1.5,k3=(0+0.25)+(1+(0.25)(1.5))=1.625,k4=(0.5)+(1+(0.5)(1.625))=2.3125,

and then

y1=1+0.56[1+2(1.5)+2(1.625)+2.3125]1.7969.

Similarly, the second step yields y23.4347.

Figure 2.6.1 presents these results together with the results (from Fig. 2.5.4) of applying the improved Euler method with step size h=0.1. We see that even with the larger step size, the Runge–Kutta method gives (for this problem) four to five times the accuracy (in terms of relative percentage errors) of the improved Euler method.

FIGURE 2.6.1.

Runge–Kutta and improved Euler results for the initial value problem dy/dx=x+y, y(0)=1.

Improved Euler Runge–Kutta
x y with h=0.1 Percent Error y with h=0.5 Percent Error Actual y
0.0 1.0000 0.00% 1.0000 0.00% 1.0000
0.5 1.7949 0.14% 1.7969 0.03% 1.7974
1.0 3.4282 0.24% 3.4347 0.05% 3.4366

It is customary to measure the computational labor involved in solving dy/dx=f(x,y) numerically by counting the number of evaluations of the function f(x, y) that are required. In Example 1, the Runge–Kutta method required eight evaluations of f(x,y)=x+y (four at each step), whereas the improved Euler method required 20 such evaluations (two for each of 10 steps). Thus the Runge–Kutta method gave over four times the accuracy with only 40% of the labor.

Computer programs implementing the Runge–Kutta method are listed in the project material for this section. Figure 2.6.2 shows the results obtained by applying the improved Euler and Runge–Kutta methods to the problem dy/dx=x+y, y(0)=1 with the same step size h=0.1. The relative error in the improved Euler value at x=1 is about 0.24%, but for the Runge–Kutta value it is 0.00012%. In this comparison the Runge–Kutta method is about 2000 times as accurate, but requires only twice as many function evaluations, as the improved Euler method.

The error bound

(8)|y(xn)yn|Ch4

for the Runge–Kutta method results in a rapid decrease in the magnitude of errors when the step size h is reduced (except for the possibility that very small step sizes may result in unacceptable roundoff errors). It follows from the inequality in (8) that (on a fixed bounded interval) halving the step size decreases the absolute error by a factor of (12)4=116. Consequently, the common practice of successively halving the step size until the computed results “stabilize” is particularly effective with the Runge–Kutta method.

FIGURE 2.6.2.

Runge–Kutta and improved Euler results for the initial value problem dy/dx=x+y, y(0)=1, with the same step size h=0.1.

x Improved Euler y Runge–Kutta y Actual y
0.1 1.1100 1.110342 1.110342
0.2 1.2421 1.242805 1.242806
0.3 1.3985 1.399717 1.399718
0.4 1.5818 1.583648 1.583649
0.5 1.7949 1.797441 1.797443
0.6 2.0409 2.044236 2.044238
0.7 2.3231 2.327503 2.327505
0.8 2.6456 2.651079 2.651082
0.9 3.0124 3.019203 3.019206
1.0 3.4282 3.436559 3.436564

Example 2

Infinite discontinuity In Example 5 of Section 2.4 we saw that Euler’s method is not adequate to approximate the solution y(x) of the initial value problem

(10)dydx=x2+y2,y(0)=1

as x approaches the infinite discontinuity near x=0.969811 (see Fig. 2.6.3). Now we apply the Runge–Kutta method to this initial value problem.

FIGURE 2.6.3.

Solutions of dy/dx=x2+y2, y(0)=1.

Figure 2.6.4 shows Runge–Kutta results on the interval [0.0, 0.9], computed with step sizes h=0.1, h=0.05, and h=0.025. There is still some difficulty near x=0.9, but it seems safe to conclude from these data that y(0.5)2.0670.

FIGURE 2.6.4.

Approximating the solution of the initial value problem in Eq. (10).

x y with h=0.1 y with h=0.05 y with h=0.025
0.1 1.1115 1.1115 1.1115
0.3 1.4397 1.4397 1.4397
0.5 2.0670 2.0670 2.0670
0.7 3.6522 3.6529 3.6529
0.9 14.0218 14.2712 14.3021

We therefore begin anew and apply the Runge–Kutta method to the initial value problem

(11)dydx=x2+y2,y(0.5)=2.0670.

Figure 2.6.5 shows results on the interval [0.5, 0.9], obtained with step sizes h=0.01, h=0.005, and h=0.0025. We now conclude that y(0.9)14.3049.

Finally, Fig. 2.6.6 shows results on the interval [0.90, 0.95] for the initial value problem

(12)dydx=x2+y2,y(0.9)=14.3049,

obtained using step sizes h=0.002, h=0.001, and h=0.0005. Our final approximate result is y(0.95)50.4723. The actual value of the solution at x=0.95 is y(0.95)50.471867. Our slight overestimate results mainly from the fact that the four-place initial value in (12) is (in effect) the result of rounding up the actual value y(0.9)14.304864; such errors are magnified considerably as we approach the vertical asymptote.

FIGURE 2.6.5.

Approximating the solution of the initial value problem in Eq. (11).

x y with h=0.01 y with h=0.005 y with h=0.0025
0.5 2.0670 2.0670 2.0670
0.6 2.6440 2.6440 2.6440
0.7 3.6529 3.6529 3.6529
0.8 5.8486 5.8486 5.8486
0.9 14.3048 14.3049 14.3049

FIGURE 2.6.6.

Approximating the solution of the initial value problem in Eq. (12).

x y with h=0.002 y with h=0.001 y with h=0.0005
0.90 14.3049 14.3049 14.3049
0.91 16.7024 16.7024 16.7024
0.92 20.0617 20.0617 20.0617
0.93 25.1073 25.1073 25.1073
0.94 33.5363 33.5363 33.5363
0.95 50.4722 50.4723 50.4723

Example 3

Skydiver A skydiver with a mass of 60 kg jumps from a helicopter hovering at an initial altitude of 5 kilometers. Assume that she falls vertically with initial velocity zero and experiences an upward force FR of air resistance given in terms of her velocity v (in meters per second) by

FR=(0.0096)(100v+10v2+v3)

(in newtons, and with the coordinate axis directed downward so that v>0 during her descent to the ground). If she does not open her parachute, what will be her terminal velocity? How fast will she be falling after 5 s have elapsed? After 10 s? After 20 s?

Solution

Newton’s law F=ma gives

mdvdt=mgFR;

that is,

(13)60dvdt=(60)(9.8)(0.0096)(100v+10v2+v3)

because m=60 and g=9.8. Thus the velocity function v(t) satisfies the initial value problem

(14)dvdt=f(v),v(0)=0,

where

(15)f(v)=9.8(0.00016)(100v+10v2+v3).

The skydiver reaches her terminal velocity when the forces of gravity and air resistance balance, so f(v)=0. We can therefore calculate her terminal velocity immediately by solving the equation

(16)f(v)=9.8(0.00016)(100v+10v2+v3)=0.

Figure 2.6.7 shows the graph of the function f(v) and exhibits the single real solution v35.5780 (found graphically or by using a calculator or computer Solve procedure). Thus the skydiver’s terminal speed is approximately 35.578 m/s, about 128 km/h (almost 80 mi/h).

FIGURE 2.6.7.

Graph of f(v)=9.8(0.00016)(100v+10v2+v3).

FIGURE 2.6.8.

The skydiver’s velocity data.

t(s) v(m/s)
0 0
1 9.636
2 18.386
3 25.299
4 29.949
5 32.678
6 34.137
7 34.875
8 35.239
9 35.415
10 35.500
11 35.541
12 35.560
13 35.569
14 35.574
15 35.576
16 35.577
17 35.578
18 35.578
19 35.578
20 35.578

Figure 2.6.8 shows the results of Runge–Kutta approximations to the solution of the initial value problem in (14); the step sizes h=0.2 and h=0.1 yield the same results (to three decimal places). Observe that the terminal velocity is effectively attained in only 15 s. But the skydiver’s velocity is 91.85% of her terminal velocity after only 5 s, and 99.78% after 10 s.

The final example of this section contains a warning: For certain types of initial value problems, the numerical methods we have discussed are not nearly so successful as in the previous examples.

Example 4

Consider the seemingly innocuous initial value problem

(17)dydx=5y6ex,y(0)=1

whose exact solution is y(x)=ex. The table in Fig. 2.6.9 shows the results obtained by applying the Runge–Kutta method on the interval [0, 4] with step sizes h=0.2, h=0.1, and h=0.05. Obviously these attempts are spectacularly unsuccessful. Although y(x)=ex0 as x+, it appears that our numerical approximations are headed toward rather than zero.

FIGURE 2.6.9.

Runge–Kutta attempts to solve numerically the initial value problem in Eq. (17).

x Runge–Kutta y with h=0.2 Runge–Kutta y with h=0.1 Runge–Kutta y with h=0.05 Actual y
0.4 0.66880 0.67020 0.67031 0.67032
0.8 0.43713 0.44833 0.44926 0.44933
1.2 0.21099 0.29376 0.30067 0.30119
1.6 0.46019 0.14697 0.19802 0.20190
2.0 4.72142 0.27026 0.10668 0.13534
2.4 35.53415 2.90419 0.12102 0.09072
2.8 261.25023 22.05352 1.50367 0.06081
3.2 1,916.69395 163.25077 11.51868 0.04076
3.6 14059.35494 1205.71249 85.38156 0.02732
4.0 103,126.5270 8903.12866 631.03934 0.01832

The explanation lies in the fact that the general solution of the equation dy/dx=5y6ex is

(18)y(x)=ex+Ce5x.

The particular solution of (17) satisfying the initial condition y(0)=1 is obtained with C=0. But any departure, however small, from the exact solution y(x)=ex—even if due only to roundoff error—introduces [in effect] a nonzero value of C in Eq. (18). And as indicated in Fig. 2.6.10, all solution curves of the form in (18) with C0 diverge rapidly away from the one with C=0, even if their initial values are close to 1.

FIGURE 2.6.10.

Direction field and solution curves for dy/dx=5y6ex.

Difficulties of the sort illustrated by Example 4 sometimes are unavoidable, but one can at least hope to recognize such a problem when it appears. Approximate values whose order of magnitude varies with changing step size are a common indicator of such instability. These difficulties are discussed in numerical analysis textbooks and are the subject of current research in the field.

2.6 Problems

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the Runge–Kutta method to approximate this solution on the interval [0, 0.5] with step size h=0.25. Construct a table showing five-decimal-place values of the approximate solution and actual solution at the points x=0.25 and 0.5.

  1. y=y, y(0)=2; y(x)=2ex

  2. y=2y, y(0)=12; y(x)=12e2x

  3. y=y+1, y(0)=1; y(x)=2ex1

  4. y=xy, y(0)=1; y(x)=2ex+x1

  5. y=yx1, y(0)=1; y(x)=2+xex

  6. y=2xy, y(0)=2; y(x)=2ex2

  7. y=3x2y, y(0)=3; y(x)=3ex3

  8. y=ey, y(0)=0; y(x)=ln (x+1)

  9. y=14(1+y2), y(0)=1; y(x)=tan14(x+π)

  10. y=2xy2, y(0)=1; y(x)=11x2

Note: The application following this problem set lists illustrative calculator/computer programs that can be used in the remaining problems.

A programmable calculator or a computer will be useful for Problems 11 through 16. In each problem find the exact solution of the given initial value problem. Then apply the Runge–Kutta method twice to approximate (to five decimal places) this solution on the given interval, first with step size h=0.2, then with step size h=0.1. Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for x an integral multiple of 0.2. Throughout, primes denote derivatives with respect to x.

  1. y=y2, y(0)=1; 0 x1

  2. y=12(y1)2, y(0)=2; 0x1

  3. yy=2x3, y(1)=3; 1x2

  4. xy=y2, y(1)=1; 1x2

  5. xy=3x2y, y(2)=3; 2x3

  6. y2y=2x5, y(2)=3; 2x3

A computer with a printer is required for Problems 17 through 24. In these initial value problems, use the Runge–Kutta method with step sizes h=0.2, 0.1, 0.05, and 0.025 to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to x.

  1. y=x2+y2, y(0)=0; 0x1

  2. y=x2y2, y(0)=1; 0x2

  3. y=x+y, y(0)=1; 0x2

  4. y=x+y3, y(0)=1; 0x2

  5. y=ln y, y(1)=2; 1x2

  6. y=x2/3+y2/3, y(0)=1; 0x2

  7. y=sin x+cos y, y(0)=0; 0x1

  8. y=x1+y2, y(1)=1; 1x1

  9. Falling parachutist As in Problem 25 of Section 2.5, you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem

    dvdt=321.6v,v(0)=0

    (with t in seconds and v in ft/s). Use the Runge–Kutta method with a programmable calculator or computer to approximate the solution for 0t2, first with step size h=0.1 and then with h=0.05, rounding off approximate v-values to three decimal places. What percentage of the limiting velocity 20 ft/s has been attained after 1 second? After 2 seconds?

  10. Deer population As in Problem 26 of Section 2.5, suppose the deer population P(t) in a small forest initially numbers 25 and satisfies the logistic equation

    dPdt=0.0225P0.0003P2

    (with t in months). Use the Runge–Kutta method with a programmable calculator or computer to approximate the solution for 10 years, first with step size h=6 and then with h=3, rounding off approximate P-values to four decimal places. What percentage of the limiting population of 75 deer has been attained after 5 years? After 10 years?

Use the Runge–Kutta method with a computer system to find the desired solution values in Problems 27 and 28. Start with step size h=1, and then use successively smaller step sizes until successive approximate solution values at x=2 agree rounded off to five decimal places.

  1. y=x2+y21, y(0)=0; y(2)=?

  2. y=x+12y2, y(2)=0; y(2)=?

Velocity-Acceleration Problems

In Problems 29 and 30, the linear acceleration a=dv/dt of a moving particle is given by a formula dv/dt=f(t,v), where the velocity v=dy/dt is the derivative of the function y=y(t) giving the position of the particle at time t. Suppose that the velocity v(t) is approximated using the Runge–Kutta method to solve numerically the initial value problem

(19)dvdt=f(t, v),v(0)=v0.

That is, starting with t0=0 and v0, the formulas in Eqs. (5) and (6) are applied—with t and v in place of x and y—to calculate the successive approximate velocity values v1, v2, v3, , vm at the successive times t1, t2, t3, , tm (with tn+1=tn+h). Now suppose that we also want to approximate the distance y(t) traveled by the particle. We can do this by beginning with the initial position y(0)=y0 and calculating

(20)yn+1=yn+vnh+12anh2

(n=1, 2, 3, ), where an=f(tn,vn)v(tn) is the particle’s approximate acceleration at time tn. The formula in (20) would give the correct increment (from yn to yn+1) if the acceleration an remained constant during the time interval [tn,tn+1].

Thus, once a table of approximate velocities has been calculated, Eq. (20) provides a simple way to calculate a table of corresponding successive positions. This process is illustrated in the project for this section, by beginning with the velocity data in Fig. 2.6.8 (Example 3) and proceeding to follow the skydiver’s position during her descent to the ground.

  1. Consider again the crossbow bolt of Example 2 in Section 2.3, shot straight upward from the ground with an initial velocity of 49 m/s. Because of linear air resistance, its velocity function v=dy/dt satisfies the initial value problem

    dvdt=(0.04)v9.8,v(0)=49

    with exact solution v(t)=294et/25245. (a) Use a calculator or computer implementation of the Runge–Kutta method to approximate v(t) for 0t10 using both n=100 and n=200 subintervals. Display the results at intervals of 1 second. Do the two approximations—each rounded to four decimal places—agree both with each other and with the exact solution? (b) Now use the velocity data from part (a) to approximate y(t) for 0t10 using n=200 subintervals. Display the results at intervals of 1 second. Do these approximate position values—each rounded to two decimal places—agree with the exact solution

    y(t)=7350(1et/25)245t?

    (c) If the exact solution were unavailable, explain how you could use the Runge–Kutta method to approximate closely the bolt’s times of ascent and descent and the maximum height it attains.

  2. Now consider again the crossbow bolt of Example 3 in Section 2.3. It still is shot straight upward from the ground with an initial velocity of 49 m/s, but because of air resistance proportional to the square of its velocity, its velocity function v(t) satisfies the initial value problem

    dvdt=(0.0011)v|v|9.8,v(0)=49.

    Beginning with this initial value problem, repeat parts (a) through (c) of Problem 29 (except that you may need n=200 subintervals to get four-place accuracy in part (a) and n=400 subintervals for two-place accuracy in part (b)). According to the results of Problems 17 and 18 in Section 2.3, the bolt’s velocity and position functions during ascent and descent are given by the following formulas.

    Ascent:v(t)=(94.388) tan(0.478837[0.103827]t),y(t)=108.465+(909.091) ln(cos(0.478837[0.103827]t));Descent:v(t)=(94.388) tanh(0.103827[t4.6119]),y(t)=108.465(909.091) ln(cosh(0.103827[t4.6119])).

2.6 Application Runge–Kutta Implementation

Figure 2.6.11 lists TI-Nspire CX CAS and Python programs implementing the Runge–Kutta method to approximate the solution of the initial value problem

dydx=x+y,y(0)=1

considered in Example 1 of this section. The comments provided in the final column should make these programs intelligible even if you have little familiarity with the Python and TI programming languages.

To apply the Runge–Kutta method to a different equation dy/dx=f(x,y), one need only change the initial line of the program, in which the function f is defined. To increase the number of steps (and thereby decrease the step size), one need only change the value of n specified in the second line of the program.

Figure 2.6.12 exhibits a Matlab implementation of the Runge–Kutta method. Suppose that the function f describing the differential equation y=f(x,y) has been defined. Then the rk function takes as input the initial value x, the initial value y, the final value x1 of x, and the desired number n of subintervals. As output it produces the resulting column vectors x and y of x- and y-values. For instance, the Matlab command

FIGURE 2.6.11.

TI-Nspire CX CAS and Python Runge–Kutta programs.

TI-Nspire CX CAS Python Comment
Define rk()=Prgm
    f(x,y):=x+y
    n:=10
    x:=0.0
    y:=1.0
    x1:=1.0
    h:=(x1-x)/n
    For i,1,n
       x0:=x
       y0:=y
       k1:=f(x,y)
       x:=x0+h/2
       y:=y0+(h*k1)/2
       k2:=f(x,y)
       y:=y0+(h*k2)/2
       k3:=f(x,y)
       x:=x0+h
       y:=y0+h*k3
       k4:=f(x,y)
       k:=(k1+2*k2+2*k3+k4)/6
       y:=y0+h*k
       Disp x,y
    EndFor
EndPrgm
# Program RK
def F(X,Y): return X + Y
N = 10
X = 0.0
Y = 1.0
X1 = 1.0
H = (X1-X)/N
for I in range(N):
X0 = X
Y0 = Y
K1 = F(X,Y)
X = X0 + H/2
Y = Y0 + H*K1/2
K2 = F(X,Y)
Y = Y0 + H*K2/2
K3 = F(X,Y)
X = X0 + H
Y = Y0 + H*K3
K4 = F(X,Y)
K = (K1+2*K2+2*K3
  +K4)/6
Y = Y0 + H*K
print (X,Y)
# END
Program title
Define function f
No. of steps
Initial x
Initial y
Final x
Step size
Begin loop
Save previous x
Save previous y
First slope
Midpoint
Midpt predictor
Second slope
Midpt predictor
Third slope
New x
Endpt predictor
Fourth slope
Average slope
Corrector
Display results
End loop
function yp = f(x,y)
yp = x + y;                          % yp = y’
function [X,Y] = rk(x,y,x1,n)
h = (x1 - x)/n;                      % step size
X = x;                               % initial x
Y = y;                               % initial y
for i = 1:n                          % begin loop
    k1 = f(x,y);                     % first slope
    k2 = f(x+h/2,y+h*k1/2);          % second slope
    k3 = f(x+h/2,y+h*k2/2);          % third slope
    k4 = f(x+h,y+h*k3);              % fourth slope
    k  = (k1+2*k2+2*k3+k4)/6;        % average slope
    x  = x + h;                      % new x
    y  = y + h*k;                    % new y
    X  = [X;x];                      % update x-column
    Y  = [Y;y];                      % update y-column
    end                              % end loop

FIGURE 2.6.12.

Matlab implementation of the Runge–Kutta method.

[X, Y] = rk(0, 1, 1, 10)

then generates the first and third columns of data shown in the table in Fig. 2.6.2.

You should begin this project by implementing the Runge–Kutta method with your own calculator or computer system. Test your program by applying it first to the initial value problem in Example 1, then to some of the problems for this section.

Famous Numbers Revisited, One Last Time

The following problems describe the numbers

e2.71828182846,ln 20.69314718056,andπ3.14159265359

as specific values of certain initial value problems. In each case, apply the Runge–Kutta method with n=10, 20, 40,  subintervals (doubling n each time). How many subintervals are needed to obtain-twice in succession-the correct value of the target number rounded to nine decimal places?

  1. The number e=y(1), where y(x) is the solution of the initial value problem dy/dx=y, y(0)=1.

  2. The number ln 2=y(2), where y(x) is the solution of the initial value problem dy/dx=1/x, y(1)=0.

  3. The number π=y(1), where y(x) is the solution of the initial value problem dy/dx=4/(1+x2), y(0)=0.

The Skydiver’s Descent

The following Matlab function describes the skydiver’s acceleration function in Example 3.

function  vp = f(t,v)
vp = 9.8 - 0.00016*(100*v + 10*v^2 + v^3);

Then the commands

k = 200                      % 200 subintervals
[t,v] = rk(0, 0, 20, k);     % Runge–Kutta approximation
[t(1:10:k+1); v(1:10:k+1)]   % Display every 10th entry

produce the table of approximate velocities shown in Fig. 2.6.8. Finally, the commands

y = zeros(k+1,1):                       % initialize y
h = 0.1;                                % step size
for n = 1:k                             % for n = 1 to k
  a = f(t(n),v(n)):                     % acceleration
  y(n+1) = y(n) + v(n)*h +0.5*a*h^2;    % Equation (20)
end                                     % end loop
[t(1:20:k+1),v(1:20:k+1),y(1:20:k+1)]   % each 20th entry

carry out the position function calculations described in Eq. (20) in the instructions for Problems 29 and 30. The results of these calculations are shown in the table in Fig. 2.6.13. It appears that the skydiver falls 629.866 m during her first 20 s of descent, and then free falls the remaining 4370.134 meters to the ground at her terminal speed of 35.578 m/s. Hence her total time of descent is 20+(4370.134/35.578)142.833 s, or about 2 min 23 s.

FIGURE 2.6.13.

The skydiver’s velocity and position data.

t (s) v (m/s) y (m)
  0 0   0  
  2 18.386 18.984
  4 29.949 68.825
  6 34.137 133.763
  8 35.239 203.392
10 35.500 274.192
12 35.560 345.266
14 35.574 416.403
16 35.577 487.555
18 35.578 558.710
20 35.578 629.866

For an individual problem to solve after implementing these methods using an available computer system, analyze your own skydive (perhaps from a different height), using your own mass m and a plausible air-resistance force of the form FR=av+bv2+cv3.

3 Linear Systems and Matrices

3.1 Introduction to Linear Systems

The subject of linear algebra centers around the problem of solving systems of (algebraic) linear equations. In high school algebra the method of elimination is used to solve systems of two and three (simultaneous) linear equations, with attention ordinarily confined to systems that have one and only one solution, and with applications to “word problems” that have one and only one “answer.” In this section we introduce some of the basic terminology of linear algebra by reviewing the elementary technique of elimination from a slightly more general viewpoint. In subsequent sections we apply this same technique to the solution of systems involving many linear equations in many unknowns. The applications of the method of elimination are diverse and important, because so many mathematical problems involve the solution of systems of linear equations.

Recall that if a, b, and c are constants with a and b not both zero, then the graph of the equation

(1)ax+by=c

is a (straight) line in the xy-plane. For this reason, an equation of the form in (1) is called a linear equation in the variables x and y. Similarly, an equation that can be written in the form

(2)ax+by+cz=d

is called linear in the three variables x, y, and z (even though its graph in xyz-space is a plane rather than a line). Thus the equations

3x2y=5andx7y+5z=11

are linear, because they involve only the first powers of the variables. By contrast, the equations

x+y+xy=5andx2+y3+z=1

are not linear, because they cannot be rewritten to eliminate the higher powers, roots, and products of the variables.

Two Equations in Two Unknowns

A system of linear equations (also called a linear system) is simply a finite collection of linear equations involving certain variables. Sometimes we refer to the variables as the “unknowns” in the system. Thus a system of two linear equations in two unknowns x and y may be written in the form

(3)a1x+b1y=c1a2x+b2y=c2.

By a solution of the system in (3) is meant a pair (x, y) of values—normally real numbers—that satisfy both equations simultaneously.

Example 1

The values x=2, y=1 constitute a solution of the system

(4)2xy=5x+2y=0

because both 2(2)(1)=5 and (2)+2(1)=0. The values x=3, y=1 satisfy the first equation in (4) but do not satisfy the second. Hence (3, 1) is not a solution of the system in (4).

Example 2

The linear system

(5)x+y=12x+2y=3

has no solution at all, because if x+y=1, then 2x+2y=2, and so 2x+2y3. Thus, two numbers that satisfy the first equation in (5) cannot simultaneously satisfy the second.

A linear system is said to be consistent if it has at least one solution and inconsistent if it has none. Thus the system in Example 1 is consistent, whereas the system in Example 2 is inconsistent.

Three Possibilities

Given a system of linear equations, we ask: What is the set of all solutions of the system? In brief, what is the solution set of the system?

In the case of a linear system

(6)a1x+b1y=c1a2x+b2y=c2

of two equations in two unknowns, we can use our knowledge of elementary geometry to sort out the possibilities for its solution set. If neither equation in (6) has both left-hand coefficients zero, then their graphs in the xy-plane are two straight lines L1 and L2. Then exactly one of the following situations must hold.

  • The lines L1 and L2 intersect at a single point (as in Figure 3.1.1).

  • The lines L1 and L2 are parallel nonintersecting lines (as in Figure 3.1.2).

  • The lines L1 and L2 coincide—they actually are the same line (see Figure 3.1.3).

A pair (x, y) of real numbers constitutes a solution of the system in (6) if and only if the point (x, y) in the coordinate plane lies on both the two lines L1 and L2. In the case shown in Figure 3.1.1, there is exactly one such point. In the case shown in Figure 3.1.2, there is no such point, and in the case in Figure 3.1.3, there are infinitely many such points—every point on the line L1=L2 is such a point. We therefore see that there are just three possibilities for a linear system of two equations in two unknowns: It has either

FIGURE 3.1.1.

Two intersecting lines: a unique solution.

FIGURE 3.1.2.

Two parallel lines: no solution.

FIGURE 3.1.3.

Two coincident lines: infinitely many solutions.

  • exactly one solution;

  • no solution; or

  • infinitely many solutions.

It is a fundamental fact of linear algebra (which we will establish in Section 3.3) that, however many equations and variables may appear in a linear system, precisely these same three possibilities occur: A system of m linear equations in n variables either has a unique solution (that is, exactly one solution), or it has no solution, or it has infinitely many solutions. Thus it is impossible, for instance, for a linear system to have exactly 2 solutions or to have exactly 17 solutions.

The Method of Elimination

The next three examples illustrate how we can use the elementary method of elimination to solve a system of two equations in two unknowns. To solve a system means to determine what its solution set is. The basic idea of the method is this:

  • First, we add an appropriate constant multiple of the first equation to the second equation. The idea is to choose the constant in such a way as to eliminate the variable x from the second equation.

  • Next, the new second equation contains only the variable y, so we readily solve it for the value of y.

  • Finally, we determine the value of x by “back substitution” of this value of y in the first equation.

Examples 3 through 5 illustrate this method in the three cases corresponding to Figures 3.1.1 through 3.1.3 (respectively).

Example 3

In order to solve the system

(7)5x+3y=1x2y=8,

we first interchange the two equations:

x2y=85x+3y=1.

Then we multiply the first equation by 5 and add the resulting terms to the second (without changing the first equation). The result is

(8)x2y=813y=39.

Now the second equation immediately yields the value y=3, and back substitution of this value in the first equation yields

x=2y+8=2(3)+8=2.

Taking it as obvious (for the moment) that the systems in (7) and (8) have the same solution set, we conclude that the original system in (7) has the unique solution x=2, y=3.

Example 4

To solve the system

(9)2x+6y=43x+9y=11,

we first multiply the first equation by 12 and get

x+3y=23x+9y=11.

We next multiply the first equation by 3 and add each term to the corresponding term in the second equation. The result is

(10)x+3y=20=5.

What, however, are we to make of the new second equation, 0=5? The system in (10) actually is

x+3y=20x+0y=5.

Because 0 is simply not equal to 5, there are no values of x and y that satisfy the second equation. Hence there certainly can be no values that satisfy both simultaneously. We conclude that the original system in (9) has no solution.

Example 5

If, instead of the system in (9), we had begun in Example 4 with the system

(11)2x+6y=43x+9y=6

and performed the same operations, we would have obtained, instead of (10), the system

(12)x+3y=20=0.

Here, 0=0 is shorthand for the equation

0x+0y=0,

which is satisfied by all values of x and y. In terms of restrictions or conditions on x and y, one of our original two equations has in effect disappeared, leaving us with the single equation

(13)x+3y=2.

Of course this is hardly surprising, because each equation in (11) is a multiple of the one in (13); in some sense we really had only one equation to begin with. At any rate, we can substitute any value of y we please in (13) and then solve for x. Thus our system in (11) has infinitely many solutions. To describe them explicitly, let us write y=t, where the parameter t is a new independent variable that we will use to generate solution pairs (x, y). Then Equation (13) yields x=23t, so our infinite solution set of the system in (11) may be described as follows:

x=23t,y=t

as the arbitrary parameter t ranges over the set of all real numbers. For instance, t=2 yields the solution (4,2), and t=3 yields the solution (11,3).

Remark

There is some latitude in how we parameterize the solutions of a system having infinitely many solutions, including the names of any parameters we introduce for this purpose. For example, we could parameterize the solutions of the system in (11) alternatively by writing x=s in (13), thereby getting the different parameterization

x=s,y=13(2s)

that yields the same solutions. For instance, the parameter values s=4 and s=11 here give the previously noted solutions (4,2) and (11,3) of the original system in (13).

Comment

These three examples illustrate the basic features of the method of elimination, which involves “transforming” a given linear system by means of a sequence of successive steps that do not change the solutions of the system. Each of these steps consists of performing one of the following three elementary operations:

  1. Multiply one equation by a nonzero constant.

  2. Interchange two equations.

  3. Add a constant multiple of (the terms of) one equation to (corresponding terms of) another equation.

In subsequent sections of this chapter we discuss the systematic use of these elementary operations to eliminate variables successively in any linear system, whatever the number of equations and variables. In this way we will see that every linear system corresponds to precisely one of the three situations illustrated in Examples 35. That is, either

  • we discover (as in Example 3) a unique solution of the system; or

  • we eventually arrive at an inconsistent equation (as in Example 4), so that the system has no solution; or

  • we find that the original system has infinitely many solutions.

Three Equations in Three Unknowns

In the case of a system of three linear equations in three variables x, y, and z, we may proceed as follows: Assuming that the first equation involves x, use the first equation to eliminate x from the second and third equations by adding appropriate multiples of the first equation. Then, assuming that the new second equation involves y, use the new second equation to eliminate y from the new third equation. Solve the new third equation for z, back-substitute in the second equation to determine y, and, finally, back-substitute y and z in the first equation to find x. The following two examples illustrate this procedure.

Example 6

Solve the linear system

(14)x+2y+z=43x+8y+7z=202x+7y+9z=23.

Solution

First, we add 3 times the first equation to the second equation; the result is

x+2y+z=42y+4z=82x+7y+9z=23.

Then addition of 2 times the first equation to the third equation yields

x+2y+z=42y+4z=83y+7z=15.

We have now eliminated x from the second and third equations. To simplify the process of eliminating y from the third equation, we multiply the second equation by 12, to obtain

x+2y+z=4y+2z=43y+7z=15.

Finally, addition of 3 times the second equation to the third equation gives

(15)x+2y+z=4y+2z=4z=3.

This system has a triangular form that makes its solution easy. By back substitution of z=3 in the second equation in (15), we find that

y=42z=42(3)=2;

then the first equation yields

x=42yz=42(2)(3)=5.

Thus our original system in (14) has the unique solution x=5, y=2, z=3.

Comment

The steps by which we transformed (14) into (15) show that every solution of the system in (14) is a solution of the system in (15). But these steps can be reversed to show similarly that every solution of the system in (15) is also a solution of the system in (14). Thus the two systems are equivalent in the sense that they have the same solution set. The computation at the end of Example 6 shows that (15) has the unique solution (x,y,z)=(5,2,3), and it follows that this also is the unique solution of the original system in (14).

Example 7

Solve the system

(16)3x8y+10z=22x3y+2z=52x9y8z=11.

Solution

In order to avoid fractions in the elimination of x, we first interchange the first two equations to get

x3y+2z=53x8y+10z=222x9y8z=11.

Addition of 3 times the first equation to the second gives

x3y+2z=5y+4z=72x9y8z=11,

and then addition of 2 times the first equation to the third equation gives

x3y+2z=5y+4z=73y12z=21.

Finally, addition of 3 times the second equation to the third equation yields

(17)x3y+2z=5y+4z=70=0.

Because the third equation has disappeared, we can choose z=t arbitrarily and then solve for x and y:

y=74z=74t,x=5+3y2z=5+3(74t)2t=2614t.

Thus our original system in (16) has infinitely many solutions. Moreover, one convenient way of describing them is this:

(18)x=2614t,y=74t,z=t.

The arbitrary parameter t can take on all real number values, and in so doing generates all the (infinitely many) solutions of the original system in (16). Figure 3.1.4 shows the three planes corresponding to the three equations in (16). These planes intersect in a straight line that is parameterized by the equations in (18).

FIGURE 3.1.4.

The three planes of Example 7.

A Differential Equations Application

In Chapter 1 we saw that a general solution of a first-order differential equation involves an arbitrary constant that must be determined in order to satisfy a given initial condition. The next example illustrates the fact that a general solution of a second-order differential equation typically involves two arbitrary constants. An initial value problem for such a differential equation ordinarily involves two initial conditions, whose imposition leads to a system of two equations with the two arbitrary constants as unknowns.

Example 8

If A and B are constants and

(19)y(x)=Ae3x+Be3x,

then differentiation yields

(20)y(x)=3Ae3x3Be3x

and

y(x)=9Ae3x+9Be3x=9y(x).

Thus the function y(x) defined by (19) satisfies the second-order differential equation

(21)y9y=0.

Now, suppose that we want to solve the initial value problem consisting of this equation and the two initial conditions

(22)y(0)=7,y(0)=9.

Then substitution of x=0 in (19) and (20) yields the linear system

A+B=73A3B=9

that we readily solve for A=5, B=2. It follows that the particular solution

y(x)=5e3x+2e3x

satisfies both the differential equation (21) and the initial conditions in (22).

In Chapter 5 we will study higher-order differential equations and initial value problems in detail, after developing the necessary methods of linear algebra in the intervening chapters.

3.1 Problems

In each of Problems 1–22, use the method of elimination to determine whether the given linear system is consistent or inconsistent. For each consistent system, find the solution if it is unique; otherwise, describe the infinite solution set in terms of an arbitrary parameter t (as in Examples 5 and 7).

  1. x+3y=92x+y=8

  2. 3x+2y=9xy=8

  3. 2x+3y=13x+5y=3

  4. 5x6y=16x5y=10

  5. x+2y=42x+4y=9

  6. 4x2y=46x3y=7

  7. x4y=102x+8y=20

  8. 3x6y=122x4y=8

  9. x+5y+z=22x+y2z=1x+7y+2z=3

  10. x+3y+2z=22x+7y+7z=12x+5y+2z=7

  11. 2x+7y+3z=11x+3y+2z=23x+7y+9z=12

  12. 3x+5yz=132x+7y+z=28x+7y+2z=32

  13. 3x+9y+7z=02x+7y+4z=02x+6y+5z=0

  14. 4x+9y+12z=13x+y+16z=462x+7y+3z=19

  15. x+3y+2z=5xy+3z=33x+y+8z=10

  16. x3y+2z=6x+4yz=45x+6y+z=20

  17. 2xy+4z=73x+2y2z=35x+y+2z=15

  18. x+5y+6z=35x+2y10z=18x+17y+8z=5

  19. x2y+z=22xy4z=13xyz=5

  20. 2x+3y+7z=15x+4y+z=20x+2y+3z=10

  21. x+yz=53x+y+3z=114x+y+5z=14

  22. 4x2y+6z=0xyz=02xy+3z=0

In each of Problems 23–30, a second-order differential equation and its general solution y(x) are given. Determine the constants A and B so as to find a solution of the differential equation that satisfies the given initial conditions involving y(0) and y(0).

  1. y+4y=0, y(x)=Acos 2x+Bsin 2x, y(0)=3, y(0)=8

  2. y9y=0, y(x)=Acosh3x+Bsinh3x, y(0)=5, y(0)=12

  3. y25y=0, y(x)=Ae5x+Be5x, y(0)=10, y(0)=20

  4. y121y=0, y(x)=Ae11x+Be11x, y(0)=44, y(0)=22

  5. y+2y15y=0, y(x)=Ae3x+Be5x, y(0)=40, y(0)=16

  6. y10y+21y=0, y(x)=Ae3x+Be7x, y(0)=15, y(0)=13

  7. 6y5y+y=0, y(x)=Aex/2+Bex/3, y(0)=7, y(0)=11

  8. 15y+y28y=0, y(x)=Ae4x/3+Be7x/5, y(0)=41, y(0)=164

  9. A system of the form

    a1x+b1y=0a2x+b2y=0,

    in which the constants on the right-hand side are all zero, is said to be homogeneous. Explain by geometric reasoning why such a system has either a unique solution or infinitely many solutions. In the former case, what is the unique solution?

  10. Consider the system

    a1x+b1y+c1z=d1a2x+b2y+c2z=d2

    of two equations in three unknowns.

    1. Use the fact that the graph of each such equation is a plane in xyz-space to explain why such a system always has either no solution or infinitely many solutions.

    2. Explain why the system must have infinitely many solutions if d1=0=d2.

  11. The linear system

    a1x+b1y=c1a2x+b2y=c2a3x+b3y=c3

    of three equations in two unknowns represents three lines L1, L2, and L3 in the xy-plane. Figure 3.1.5 shows six possible configurations of these three lines. In each case describe the solution set of the system.

  12. Consider the linear system

    a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3

    of three equations in three unknowns to represent three planes P1, P2, and P3 in xyz-space. Describe the solution set of the system in each of the following cases.

    1. The three planes are parallel and distinct.

    2. The three planes coincide—P1=P2=P3.

    3. P1 and P2 coincide and are parallel to P3.

    4. P1 and P2 intersect in a line L that is parallel to P3.

    5. P1 and P2 intersect in a line L that lies in P3.

    6. P1 and P2 intersect in a line L that intersects P3 in a single point.

FIGURE 3.1.5.

Three lines in the plane (Problem 33).

3.2 Matrices and Gaussian Elimination

In Example 6 of Section 3.1 we applied the method of elimination to solve the linear system

(1)1x+2y+1z=43x+8y+7z=202x+7y+9z=23.

There we employed elementary operations to transform this system into the equivalent system

(2)1x+2y+1z=40x+1y+2z=40x+0y+1z=3,

which we found easy to solve by back substitution. Here we have printed in color the coefficients and constants (including the 0s and 1s that would normally be omitted) because everything else—the symbols x, y, and z for the variables and the + and = signs—is excess baggage that means only extra writing, for we can keep track of these symbols mentally. In effect, in Example 6 we used an appropriate sequence of operations to transform the array

(3)[12143872027923]

of coefficients and constants in (1) into the array

(4)[121401240013]

of constants and coefficients in (2).

Rectangular arrays of numbers like those in (3) and (4) are called matrices. Thus, a matrix is simply a rectangular array of numbers, which are called the entries or elements of the matrix. The size, or shape, of a matrix is specified by telling how many horizontal rows and vertical columns it has. Each matrix in (3) and (4) has three rows and four columns; therefore, each is a 3×4 (read “three by four”) matrix. We always specify first the number of rows and then the number of columns.

Example 1

The matrices

[3725][3015][203]

have sizes 2×2, 1×4, and 3×1, respectively.

Coefficient Matrices

A general system of m linear equations in the n variables x1,x2,,xn may be written in the form

(5)a11x1+a12x2+a13x3++a1nxn=b1a21x1+a22x2+a23x3++a2nxn=b2am1x1+am2x2+am3x3++amnxn=bm.

Observe that aij denotes the (constant) coefficient in the ith equation of the jth variable xj and that bi denotes the constant on the right-hand side in the ith equation. Thus the first subscript i refers to the equation and the second subscript j to the variable. For instance, a32 denotes the coefficient of x2 in the 3rd equation. This scheme of systematic double subscripts enables us to specify readily the location of each coefficient in a system with many equations and variables. It also helps us get rid of the excess baggage in (5) by focusing our attention on the coefficients.

The coefficient matrix of the linear system in (5) is the m×n matrix

(6)A=[a11a12a13a1na21a22a23a2nam1am2am3amn].

We will use boldface capital letters to denote matrices and lowercase letters to denote numbers. When we want to refer more briefly to the matrix A and its entries, we can write

(7)A=[aij]=[aij](jth column)(ith row)

The first subscript i specifies the row and the second subscript j the column of A in which the element aij appears:

First subscript Second subscript
Row Column

Although double subscripts may seem tedious when first encountered, their usefulness should not be underestimated. For instance, they are consistent with the notation used in most programming languages. Many practical applications lead to linear systems with hundreds or even thousands of variables and equations. In a typical computer system, the coefficient matrix A of such a system would be represented by a two-dimensional array in which A(i, j) denotes aij. With this approach, the computer can store a 100×100 matrix in the same way that it stores a 3×3 matrix.

The matrices in (3) and (4) include not only the coefficients but also the constants on the right-hand sides in the corresponding linear systems in (1) and (2). Let us write

(8)b=[b1b2bm]

for the column of constants in the general system in (5). An m×1 matrix—that is, one with a single column—is often called a (column) vector and is denoted by a boldface letter. When we adjoin the constant vector b to the coefficient matrix A (as a final column), we get the matrix

(9)[Ab]=[a11a12a1nb1a21a22a2nb2am1am2amnbm].

This m×(n+1) matrix is called the augmented coefficient matrix, or simply the augmented matrix, of the m×n system in (5).

Although it takes a good deal of notation to describe the augmented matrix of a general linear system, it is a very simple matter to write the augmented matrix of a particular system. If the system is written in chalk on a blackboard with the variables in the same order in each equation, we merely erase all the xj’s and the plus and equal signs (retaining a minus sign for each negative coefficient) and insert a zero in each spot where a variable is missing in an equation.

Example 2

The augmented coefficient matrix of the system

2x1+3x27x3+4x4=6x2+3x35x4=0x1+2x29x4=17

of three equations in four variables is the 3×5 matrix

[2374601350120917].

Elementary Row Operations

In Section 3.1 we described the three elementary operations that are used in the method of elimination. To each of these corresponds an elementary row operation on the augmented matrix of the system. For instance, when we interchange two equations in the system, we interchange the corresponding two rows of its augmented coefficient matrix.

Figure 3.2.1 shows notation that we use to describe elementary row operations briefly. For instance,

(10)[1234](3)R1+R2[12610]

shows the result of adding 3 times row 1 to row 2. Note that, when we perform the operation (c)Rp+Rq on a matrix—that is, when we add c times row p to row q—row p itself remains unchanged.

FIGURE 3.2.1.

Notation for elementary row operations.

Type Row Operation Notation
1 Multiply row p by c cRp
2 Interchange row p and row q SWAP(Rp,Rq)
3 Add c times row p to row q (c)Rp+Rq

Example 3

To solve the system

(11)x1+2x2+x3=43x1+8x2+7x3=202x1+7x2+9x3=23,

whose augmented coefficient matrix is exhibited in (3), we carry out the following sequence of elementary row operations, corresponding to the steps in the solution of Example 6 in Section 3.1:

(12)
(13)
[12143872027923](3)R1+R2[1214024827923](2)R1+R3[1214024803715](12)R2[1214012403715](3)R2+R3[121401240013].

The final matrix here is the augmented coefficient matrix of the system

(14)x1+2x2+x3=4x2+2x3=4.x3=3

whose unique solution (readily found by back substitution) is x1=5, x2=2, x3=3.

It is not quite self-evident that a sequence of elementary row operations produces a linear system having the same solution set as the original system. To state the pertinent result concisely, we need the following definition.

Thus the two matrices in (10) are row equivalent, as are the two matrices in (12) and (13). In Problem 29 we ask you to show that if B can be obtained from A by elementary row operations, then A can be obtained from B by elementary row operations. This follows from the observation that row operations are “reversible.” In Problem 30 we suggest a proof of the following theorem.

Thus, the linear systems in (11) and (14) have the same solution set, because their augmented matrices in (12) and (13) are row equivalent.

Echelon Matrices

Up to this point we have been somewhat informal in our description of the method of elimination. Its objective is to transform a given linear system, by elementary row operations, into one for which back substitution leads easily and routinely to a solution. The following definition tells what should be the appearance of the augmented coefficient matrix of the transformed system.

Echelon matrices are sometimes called row-echelon matrices. Property 1 says that if E has any all-zero rows, then they are grouped together at the bottom of the matrix. The first (from the left) nonzero element in each of the other rows is called its leading entry. Property 2 says that the leading entries form a “descending staircase” pattern from upper left to lower right, as in the following echelon matrix.

Here we have highlighted the leading entries and indicated the descending staircase pattern.

The following list exhibits all possibilities for the form of a 3×3 echelon matrix, with asterisks denoting the (nonzero) leading entries; a # denotes an element that may be either zero or nonzero.

[##0#00][000000000][##0#000][##00000][0#00000][##000000][0#000000][00000000]

It follows from Properties 1 and 2 that the elements beneath any leading entry in the same column are all zero. The matrices

A=[132000015]andB=[015132000]

are not echelon matrices because A does not have Property 1 and B does not have Property 2.

Suppose that a linear system is in echelon form—its augmented matrix is an echelon matrix. Then those variables that correspond to columns containing leading entries are called leading variables; all the other variables are called free variables. The following algorithm describes the process of back substitution to solve such a system.

Example 4

The augmented coefficient matrix of the system

(15)x12x2+3x3+2x4+x5=10x3+2x5=3x44x5=7

is the echelon matrix

(16)[1232110001023000147].

The leading entries are in the first, third, and fourth columns. Hence x1, x3, and x4 are the leading variables and x2 and x5 are the free variables. To solve the system by back substitution, we first write

(17a)x2=s,x5=t,

where s and t are arbitrary parameters. Then the third equation in (15) gives

(17b)x4=7+4x5=7+4t;

the second equation gives

(17c)x3=32x5=32t;

finally, the first equation in (15) yields

x1=10+2x23x32x4x5=10+2s3(32t)2(7+4t)t.

Therefore,

(17d)x1=5+2s3t.

Thus the system in (15) has an infinite solution set consisting of all (x1,x2,x3,x4,x5) given in terms of the two parameters s and t as follows:

x1=5+2s3tx2=sx3=32tx4=7+4tx5=t.

For instance, with s=2 and t=1 we get the solution x1=6, x2=2, x3=5, x4=11, and x5=1.

Gaussian Elimination

Because we can use back substitution to solve any linear system already in echelon form, it remains only to establish that we can transform any matrix (using elementary row operations) into an echelon matrix. The procedure that we have already illustrated in several examples is known as Gaussian elimination. The following systematic description of the procedure makes it clear that it succeeds with any given matrix A.

In brief, we work on the matrix A one column at a time, from left to right. In each column containing a leading entry (perhaps after a row interchange), we “clear out” the nonzero elements below it and then move on to the next column.

Example 5

To solve the system

(18)x12x2+3x3+2x4+x5=102x14x2+8x3+3x4+10x5=73x16x2+10x3+6x4+5x5=27,

we reduce its augmented coefficient matrix to echelon form as follows.

[1232110248310736106527](2)R1+R2[1232110002181336106527](3)R1+R3[12321100021813001023]SWAP(R2,R3)[12321100010230021813](2)R2+R3[1232110001023000147](1)R3[1232110001023000147]

Our final result is the echelon matrix in (16), so by Eqs. (17a)(17d) in Example 4, the infinite solution set of the system in (18) is described in terms of arbitrary parameters s and t as follows:

(19)x1=5+2s3tx2=sx3=32tx4=7+4tx5=t.

Thus, the substitution of any two specific values for s and t in (19) yields a particular solution (x1,x2,x3,x4,x5) of the system, and each of the system’s infinitely many different solutions is the result of some such substitution.

Examples 3 and 5 illustrate the ways in which Gaussian elimination can result in either a unique solution or infinitely many solutions. On the other hand, if the reduction of the augmented matrix to echelon form leads to a row of the form

0000,

where the asterisk denotes a nonzero entry in the last column, then we have an inconsistent equation,

0x1+0x2++0xn=*,

and therefore the system has no solution.

Remark

We use algorithms such as the back substitution and Gaussian elimination algorithms of this section to outline the basic computational procedures of linear algebra. In modern numerical work, these procedures often are implemented on a computer. For instance, linear systems of more than four equations are usually solved in practice by using a computer to carry out the process of Gaussian elimination.

3.2 Problems

The linear systems in Problems 1–10 are in echelon form. Solve each by back substitution.

  1. x1+x2+2x3=5x2+3x3=6x3=2

  2. 2x15x2+x3=23x22x3=9x3=3

  3. x13x2+4x3=7x25x3=2

  4. x15x2+2x3=10x27x3=5

  5. x1+x22x3+x4=9x2x3+2x4=1x33x4=5

  6. x12x2+5x33x4=7x23x3+2x4=3x4=4

  7. x1+2x2+4x35x4=17x22x3+7x4=7

  8. x110x2+3x313x4=5x3+3x4=10

  9. 2x1+x2+x3+x4=63x2x32x4=23x3+4x4=9x4=6

  10. x15x2+2x37x4+11x5=0x213x3+3x47x5=0x45x5=0

In Problems 11–22, use elementary row operations to transform each augmented coefficient matrix to echelon form. Then solve the system by back substitution.

  1. 2x1+8x2+3x3=2x1+3x2+2x3=52x1+7x2+4x3=8

  2. 3x1+x23x3=62x1+7x2+x3=92x1+5x2=5

  3. x1+3x2+3x3=132x1+5x2+4x3=232x1+7x2+8x3=29

  4. 3x16x22x3=12x14x2+x3=17x1+2x22x3=9

  5. 3x1+x23x3=4x1+x2+x3=15x1+6x2+8x3=8

  6. 2x1+5x2+12x3=63x1+x2+5x3=125x1+8x2+21x3=17

  7. x14x23x33x4=42x16x25x35x4=53x1x24x35x4=7

  8. 3x16x2+x3+13x4=153x16x2+3x3+21x4=212x14x2+5x3+26x4=23

  9. 3x1+x2+x3+6x4=14x12x2+5x35x4=74x1+x2+2x3+7x4=17

  10. 2x1+4x2x32x4+2x5=6x1+3x2+2x37x4+3x5=95x1+8x27x3+6x4+x5=4

  11. x1+x2+x3=62x12x25x3=133x1+x3+x4=134x12x23x3+x4=1

  12. 4x12x23x3+x4=32x12x25x3=104x1+x2+2x3+x4=173x1+x3+x4=12

In Problems 23–27, determine for what values of k each system has (a) a unique solution; (b) no solution; (c) infinitely many solutions.

  1. 3x+2y=16x+4y=k

  2. 3x+2y=06x+ky=0

  3. 3x+2y=116x+ky=21

  4. 3x+2y=17x+5y=k

  5. x+2y+z=32xy3z=54x+3yz=k

  6. Under what condition on the constants a, b, and c does the system

    2xy+3z=ax+2y+z=b7x+4y+9z=c

    have a unique solution? No solution? Infinitely many solutions?

  7. This problem deals with the reversibility of elementary row operations.

    1. If the elementary row operation cRp changes the matrix A to the matrix B, show that (1/c)Rp changes B to A.

    2. If SWAP(Rp,Rq) changes A to B, show that SWAP(Rp,Rq) also changes B to A.

    3. If cRp+Rq changes A to B, show that (c)Rp+Rq changes B to A.

    4. Conclude that if A can be transformed into B by a finite sequence of elementary row operations, then B can similarly be transformed into A.

  8. This problem outlines a proof that two linear systems LS1 and LS2 are equivalent (that is, have the same solution set) if their augmented coefficient matrices A1 and A2 are row equivalent.

    1. If a single elementary row operation transforms A1 to A2, show directly—considering separately the three cases—that every solution of LS1 is also a solution of LS2.

    2. Explain why it now follows from Problem 29 that every solution of either system is also a solution of the other system; thus the two systems have the same solution set.

3.2 Application Automated Row Reduction

Computer algebra systems are often used to ease the labor of matrix computations, including elementary row operations. The 3×4 augmented coefficient matrix of Example 3 can be entered with the Maple command

with(linalg):
A := array( [[1, 2, 1,  4],
             [3, 8, 7, 20],
             [2, 7, 9, 23]] );

or the Mathematica command


A = {{1, 2, 1,  4},
     {3, 8, 7, 20},
     {2, 7, 9, 23}}

or the Matlab command


A = [1 2 1  4
     3 8 7 20
     2 7 9 23]

The Maple linalg package has built-in elementary row operations that can be used to carry out the reduction of A exhibited in Example 3, as follows:


A := addrow(A,1,2,-3);          # (-3)R1 + R2
A := addrow(A,1,3,-2);          # (-2)R1 + R3
A := mulrow(A,2,1/2);           # (1/2)R2
A := addrow(A,2,3,-3);          # (-3)R2 + R3

The inserted remarks, which follow the notation of Fig. 3.2.1, should make clear the structure of these row operations in Maple.

In Mathematica the ith row of the matrix A is denoted by A[[i]], and we can operate directly on rows to carry out the reduction of Example 3, as follows:


A[[2]] = (-3)A[[1]] + A[[2]];   (* (-3)R1 + R2  *)
A[[3]] = (-2)A[[1]] + A[[3]];   (* (-2)R1 + R3  *)
A[[2]] = (1/2)A[[2]];           (*  (1/2)R2     *)
A[[3]] = (-3)A[[2]] + A[[3]];   (* (-3)R2 + R3  *)

In Matlab, the ith row of the matrix A is denoted by A(i,:), and we can operate similarly on rows to carry out the reduction to echelon form, as follows:


A(2,:) = (-3)*A(1,:)   + A(2,:)  %  (-3)R1  + R2
A(3,:) = (-2)*A(1,:)   + A(3,:)  %  (-2)R1  + R3
A(2,:) = (1/2)*A(2,:)            %  (1/2)R2
A(3,:) = (-3)*A(2,:)   + A(3,:)  %  (-3)R2  + R3

You should verify (by using the computer algebra system of your choice) that these operations yield the echelon matrix

[121401240013].

If you wanted to interchange the 1st and 3rd rows—and proceed to solve the corresponding system by “forward substitution” rather than back substitution—this could be done as follows:


A := swaprow(A,1,3);
A[[{1,3}]] = A[[{3,1}]];
A([1 3],:)  = A([3 1],:)

(Maple)
(Mathematica)
(MATLAB)

These “automated” elementary row operations can be used in Problems 11–22 of this section.

3.3 Reduced Row-Echelon Matrices

The result of the process of Gaussian elimination described in Section 3.2 is not uniquely determined. That is, two different sequences of elementary row operations, both starting with the same matrix A, may yield two different echelon matrices, each of course still row equivalent to A. For instance, the two echelon matrices

(1)[123045006]and[111022003]

are readily seen (as in Problem 31) to be row equivalent. Hence it is possible to begin with an appropriate 3×3 matrix and derive by Gaussian elimination either of the two different echelon matrices in (1).

A full understanding of the structure of systems of linear equations depends on the definition of a special class of echelon matrices, a class having the property that every matrix in it is row equivalent to one and only one of these special echelon matrices. Recall that an echelon matrix E is one that has the following two properties:

  1. Every all-zero row of E lies beneath every row that contains a nonzero element.

  2. The leading nonzero entry in each row lies to the right of the leading nonzero entry in the preceding row.

A reduced echelon matrix (sometimes called a reduced row-echelon matrix) has two additional properties.

A matrix is said to be in reduced echelon form if it is a reduced echelon matrix. Similarly, a linear system is in reduced echelon form if its augmented coefficient matrix is a reduced echelon matrix.

Example 1

The following matrices are reduced echelon matrices.

[1001][103001400001][120001000][107015000]

The echelon matrices

A=[100020000]andB=[102010001]

are not in reduced echelon form, because A does not have Property 3 and B does not have Property 4.

The process of transforming a matrix A into reduced echelon form is called Gauss-Jordan elimination.

The reduced echelon form of a matrix is unique. The special class of matrices that we mentioned at the beginning of this discussion is simply the class of all reduced echelon matrices. A proof of the following theorem may be found in Section 4.2 of B. Noble and J. W. Daniel, Applied Linear Algebra, 3rd ed., Hoboken, NJ: Pearson, 1988.

Example 2

Find the reduced echelon form of the matrix

A=[12143872027923].

Solution

In Example 3 of Section 3.2 we found the echelon form

[121401240013],

which already satisfies Property 3. To clear out columns 2 and 3 (in order to satisfy Property 4), we continue the reduction as follows.

[121401240013](2)R2+R1[103401240013](2)R3+R2[103401020013](3)R3+R1[100501020013]

For instance, we see immediately from this reduced echelon form that the linear system

x+2y+z=43x+8y+7z=202x+7y+9z=23

with augmented coefficient matrix A has the unique solution x=5, y=2, z=3.

Example 3

To use Gauss-Jordan elimination to solve the linear system

(2)x1+x2+x3+x4=12x1+2x2+5x4=173x1+2x2+4x3x4=31,

we transform its augmented coefficient matrix into reduced echelon form, as follows:

[111112120517324131](1)R1+R2[11111201145324131](3)R1+R3[1111120114501145](1)R2+R3[1111120114500000](1)R2+R1[102370114500000].

Thus the reduced echelon form of the system in (2) is

(3)x1+2x33x4=7x2x3+4x4=50=0.

The leading variables are x1 and x2; the free variables are x3 and x4. If we set

x3=sandx4=t,

then (3) immediately yields

x1=72s+3t,x2=5+s4t.

As a practical matter, Gauss-Jordan elimination generally offers no significant computational advantage over Gaussian elimination (transformation to nonreduced echelon form) followed by back substitution. Therefore, Gaussian elimination is commonly employed in practical procedures and in computer programs used to solve linear systems numerically.

The Three Possibilities

The chief importance of Gauss-Jordan elimination stems from the fact that the reduced echelon form of a general linear system

(4)a11x1+a12x2+a13x3++a1nxn=b1a21x1+a22x2+a23x3++a2bxn=b2am1x1+am2x2+am3x3++amnxn=bm

most clearly exhibits its underlying structure and enables us most readily to answer questions about the number and type of its solutions. If the leading variables are xj1,xj2,,xjr, then the reduced echelon form of the system in (4) looks like

(5)xj1+k=r+1nc1kxjk=d1xj2+k=r+1nc2kxjk=d2xjr+k=r+1ncrkxjk=dr0=dr+10=dm,

where the summations involve only the (non-leading) free variables xjr+1,,xjn.

Now if any of the constants dr+1,dr+2,,dm in (5) is nonzero, then clearly the system has no real solution. On the other hand, suppose that dr+1=dr+2==dm=0. If r<n, then there are free variables that we can set equal to arbitrary parameters, and so the system has infinitely many solutions. If instead r=n, then the summations in (5) disappear and we have the unique solution x1=d1, x2=d2, , xn=dn. These observations establish the following result, to which we have already alluded.

Homogeneous Systems

The linear system in (4) is called homogeneous provided that the constants b1,b2, ,bm on the right-hand side are all zero. Thus a homogeneous system of m equations in n variables has the form

(6)a11x1+a12x2+a13x3++a1nxn=0a21x1+a22x2+a23x3++a2nxn=0am1x1+am2x2+am3x3++amnxn=0.

Every homogeneous system obviously has at least the trivial solution

(7)x1=0,x2=0,,xn=0.

Thus (by Theorem 2) we know from the outset that every homogeneous linear system either has only the trivial solution or has infinitely many solutions. If it has a nontrivial solution—one with not every xi equal to zero—then it must have infinitely many solutions.

An important special case, in which a nontrivial solution is guaranteed, is that of a homogeneous system with more variables than equations: m<n. To see why there must be a solution, consider the reduced echelon system in (5) with the constants on the right-hand side all zero (because the original system is homogeneous). The number r of leading variables is at most the number m of equations (because there is at most one leading variable per equation). If m<n, it then follows that r<n, so there is at least one free variable that can be set equal to an arbitrary parameter, thereby yielding infinitely many solutions. This argument establishes the following key result.

Example 4

The homogeneous linear system

47x173x2+56x3+21x4=019x1+81x217x399x4=053x1+62x2+39x3+25x4=0

of three equations in four unknowns necessarily has infinitely many solutions. The only question (which we could answer by reducing the system to echelon form) is whether the system has one, two, or three free variables.

The situation is different for a nonhomogeneous system with more variables than equations. The simple example

x1+x2+x3=0x1+x2+x3=1

shows that such a system may be inconsistent. But if it is consistent—meaning that dr+1==dm=0 in the reduced echelon form in (5)—then the fact that m<n implies (just as in the proof of Theorem 3) that there is at least one free variable, and that the system therefore has infinitely many solutions. Hence every nonhomogeneous system with more variables than equations either has no solution or has infinitely many solutions.

Equal Numbers of Equations and Variables

An especially important case in the theory of linear systems is that of a homogeneous system

(8)a11x1+a12x2+a13x3++a1nxn=0a21x1+a22x2+a23x3++a2nxn=0an1x1+an2x2+an3x3++annxn=0

with the same number n of variables and equations. The coefficient matrix A=[aij] then has the same number of rows and columns and thus is an n×n square matrix.

Here we are most interested in the situation when (8) has only the trivial solution x1=x2==xn=0. This occurs if and only if the reduced echelon system contains no free variables. That is, all n of the variables x1,x2,,xn must be leading variables. Because the system consists of exactly n equations, we conclude that the reduced echelon system is simply

x1=0x2=0x3=0xn=0,

and, therefore, that the reduced echelon form of the coefficient matrix A is the matrix

(9)[1000010000100001].

Such a (square) matrix, with ones on its principal diagonal (the one from upper left to lower right) and zeros elsewhere, is called an identity matrix (for reasons given in Section 3.4). For instance, the 2×2 and 3×3 identity matrices are

[1001]and[100010001].

The matrix in (9) is the n×n identity matrix. With this terminology, the preceding argument establishes the following theorem.

Example 5

The computation in Example 2 (disregarding the fourth column in each matrix there) shows that the matrix

A=[121387279]

is row equivalent to the 3×3 identity matrix. Hence Theorem 4 implies that the homogeneous system

x1+2x2+x3=03x1+8x2+7x3=02x1+7x2+9x3=0

with coefficient matrix A has only the trivial solution x1=x2=x3=0.

3.3 Problems

Find the reduced echelon form of each of the matrices given in Problems 1–20.

  1. [1237]

  2. [3725]

  3. [37152511]

  4. [371528]

  5. [12112319]

  6. [12194770]

  7. [123141219]

  8. [145393123]

  9. [52180144112]

  10. [525947417]

  11. [391267136]

  12. [1423121285]

  13. [274013212654]

  14. [1325252327722]

  15. [2242114327193]

  16. [1315724228273417]

  17. [1111412281231311]

  18. [125121231811925262111]

  19. [271019131348610213]

  20. [36171351081847245926]

  1. 21–30. Use the method of Gauss-Jordan elimination (transforming the augmented matrix into reduced echelon form) to solve Problems 11–20 in Section 3.2 .

  1. Show that the two matrices in (1) are both row equivalent to the 3×3 identity matrix (and hence, by Theorem 1, to each other).

  2. Show that the 2×2 matrix

    A=[abcd]

    is row equivalent to the 2×2 identity matrix provided that adbc0.

  3. List all possible reduced row-echelon forms of a 2×2 matrix, using asterisks to indicate elements that may be either zero or nonzero.

  4. List all possible reduced row-echelon forms of a 3×3 matrix, using asterisks to indicate elements that may be either zero or nonzero.

  5. Consider the homogeneous system

    ax+by=0cx+dy=0.
    1. If x=x0 and y=y0 is a solution and k is a real number, then show that x=kx0 and y=ky0 is also a solution.

    2. If x=x1, y=y1 and x=x2, y=y2 are both solutions, then show that x=x1+x2, y=y1+y2 is a solution.

  6. Suppose that adbc0 in the homogeneous system of Problem 35. Use Problem 32 to show that its only solution is the trivial solution.

  7. Show that the homogeneous system in Problem 35 has a nontrivial solution if and only if adbc=0.

  8. Use the result of Problem 37 to find all values of c for which the homogeneous system

    (c+2)x+3y=02x+(c3)y=0

    has a nontrivial solution.

  9. Consider a homogeneous system of three equations in three unknowns. Suppose that the third equation is the sum of some multiple of the first equation and some multiple of the second equation. Show that the system has a nontrivial solution.

  10. Let E be an echelon matrix that is row equivalent to the matrix A. Show that E has the same number of nonzero rows as does the reduced echelon form E* of A. Thus the number of nonzero rows in an echelon form of A is an “invariant” of the matrix A. Suggestion: Consider reducing E to E*.

3.3 Application Automated Row Reduction

Most computer algebra systems include commands for the immediate reduction of matrices to reduced echelon form. For instance, if the matrix

A=[12143872027923]

of Example 2 has been entered—as illustrated in the 3.2 Application—then the Maple command


with(linalg):  R := rref(A);

or the Mathematica command


R = RowReduce[A] // MatrixForm

or the Matlab command


R = rref(A)

or the Wolfram|Alpha query


row reduce ((1, 2, 1, 4), (3, 8, 7, 20), (2, 7, 9, 23))

produces the reduced echelon matrix

R=[100501020013]

that exhibits the solution of the linear system having augmented coefficient matrix A. The same calculation is illustrated in the calculator screen of Fig. 3.3.1. Solve similarly the systems in the following problems (which apparently would be somewhat tedious to solve by manual row reduction):

FIGURE 3.3.1.

Finding a reduced row-echelon matrix with a TI-89 calculator.

  1. 17x+42y36z=21313x+45y34z=22612x+47y35z=197

  2. 32x+57y41z=71323x+43y37z=13042x61y+39z=221

  3. 231x+157y241z=420323x+181y375z=412542x+161y759z=419

  4. 837x+667y729z=1659152x179y975z=1630542x+328y759z=1645

  5. 49w57x+37y59z=9773w15x19y22z=9952w51x+14y29z=8913w27x+27y25z=73

  6. 64w57x+97y67z=48592w+77x34y37z=48644w34x+53y34z=46527w+57x69y+29z=464

3.4 Matrix Operations

As yet, we have used matrices only to simplify our record keeping in the solution of linear systems. But it turns out that matrices can be added and multiplied in ways similar to the ways in which numbers are added and multiplied and that these operations with matrices have far-reaching applications.

At the level of this text everyone “knows” that 2+3=5, and we do not dwell on the underlying meaning of this equation. But in the case of matrices we must begin with precise definitions of what the familiar language of algebra is to mean when it is applied to matrices rather than to numbers.

Two matrices A and B of the same size—the same number of rows and the same number of columns—are called equal provided that each element of A is equal to the corresponding element of B. Thus two matrices of the same size are equal provided they are elementwise equal, and we write A=B to denote equality of the two matrices A and B.

Example 1

If

A=[3456],B=[3457],andC=[347568],

then AB because a22=6, whereas b22=7, and AC because the matrices A and C are not of the same size.

The next two definitions are further examples of “doing it elementwise.”

Example 2

If

A=[301275],B=[436902],andC=[3216],

then

A+B=[7351173],

but the sum A+C is not defined because the matrices A and C are not of the same size.

Using multiplication of a matrix by a scalar, we define the negative A of the matrix A and the difference AB of the two matrices A and B by writing

A=(1)AandAB=A+(B).

Example 3

If A and B are the 2×3 matrices of Example 2, then

3A=[90362115],B=[436902],

and

3AB=[53932117].

Vectors

Our first application of these matrix operations is to vectors. As mentioned in Section 3.2, a column vector (or simply vector) is merely an n×1 matrix, one having a single column. We normally use boldface lowercase letters, rather than lightface uppercase letters, to denote vectors. If

a=[625]andb=[234],

then we can form such combinations as

3a+2b=[18615]+[468]=[1407].

Largely for typographical reasons, we sometimes write

(3)a=[a1a2an]=(a1,a2,,an).

That is, (a1,a2,,an) is simply another notation for the column vector with elements a1,a2,,an. It should not be confused with the row vector

(4)[a1a2an].

A row vector is a 1×n (rather than n×1) matrix having a single row, and

(3, 2, 1)=[321][321]

because the two matrices here have different sizes (even though they have the same elements).

Now consider the linear system

(5)a11x1+a12x2+a13x3++a1nxn=b1a21x1+a22x2+a23x3++a2nxn=b2am1x1+am2x2+am3x3++amnxn=bm

of m equations in n variables. We may regard a solution of this system as a vector

(6)x=[x1x2x3xn]=(x1,x2,x3,,xn)

whose elements satisfy each of the equations in (5). If we want to refer explicitly to the number of elements, we may call x an n-vector.

Example 4

Consider the homogeneous system

(7)x1+3x215x3+7x4=0x1+4x219x3+10x4=02x1+5x226x3+11x4=0.

We find readily that the reduced echelon form of the augmented coefficient matrix of this system is

[103200143000000].

Hence x1 and x2 are leading variables and x3 and x4 are free variables. In the manner of Sections 3.2 and 3.3, we therefore see that the infinite solution set of the system in (7) is described by the equations

(8)x4=t,x3=s,x2=4s3t,x1=3s+2t

in terms of the arbitrary parameters s and t.

Now let us write the solution x=(x1,x2,x3,x4) in vector notation. The equations in (8) yield

x=[x1x2x3x4]=[3s+2t4s3tst],

and “separating” the s and t parts gives

x=[3s4ss0]+[2t3t0t]=s[3410]+t[2301]

—that is,

(9)x=s(3,4,1,0)+t(2,3,0,1)=sx1+tx2.

Equation (9) expresses in vector form the general solution of the linear system in (7). It says that the vector x is a solution if and only if x is a linear combination—a sum of multiples—of the particular solutions x1=(3,4,1,0) and x2=(2,3,0,1). The parameters s and t are simply the coefficients in this “sum of multiples.”

In the same manner as that in which we derived Eq. (9) from the equations in (8), the general solution of every homogeneous linear system can be expressed as a linear combination of particular solution vectors. For this reason (as well as others), linear combinations of vectors will play a central role in succeeding chapters.

Matrix Multiplication

The first surprise is that matrices are not multiplied elementwise. The initial purpose of matrix multiplication is to simplify the notation for systems of linear equations. If we write

(10)A=[aij],x=[x1x2xn],andb=[b1b2bm],

then A, x, and b are, respectively, the coefficient matrix, the unknown vector, and the constant vector for the linear system in (5). We want to define the matrix product Ax in such a way that the entire system of linear equations reduces to the single matrix equation

(11)Ax=b.

The first step is to define the product of a row vector a and a column vector b,

a=[a1a2an]andb=[b1b2bn],

each having n elements. In this case, the product ab is defined to be

(12)ab=a1b1+a2b2++anbn.

Thus ab is the sum of products of corresponding elements of a and b. For instance,

[23][35]=(2)(3)+(3)(5)=9

and

[3017][5234]=35+02+(1)(3)+74=46.

Note that if

a=[a1a2an]andx=[x1x2xn],

then

ax=a1x1+a2x2++anxn.

Hence the single equation

(13)a1x1+a2x2++anxn=b

reduces to the equation

(14)ax=b,

which is a step toward the objective expressed in Eq. (11). This observation is the underlying motivation for the following definition.

That is, if the ith row of A is

[ai1ai2ai3aip]

and the jth column of B is

[b1jb2jb3jbpj],

then the element in the ith row and jth column of the product AB is

ai1b1j+ai2b2j+ai3b3j++aipbpj.

Example 5

If

A=[2143]andB=[1537],

then m=p=n=2, so AB will also be a 2×2 matrix. To find AB, we calculate sums of products as follows:

AB, row 1, column 1:(2)(1)+(1)(3)=1;AB, row 1, column 2:(2)(5)+(1)(7)=3;AB, row 2, column 1:(4)(1)+(3)(3)=5;AB, row 2, column 2:(4)(5)+(3)(7)=1.

Thus

AB=[1351].

For your first practice with matrix multiplication, you should compute

BA=[1537][2143]=[18142218].

Note that ABBA. This shows that multiplication of matrices is not commutative! We must therefore be careful about the order in which we write the matrices in a matrix product.

The definition of the matrix product bears careful examination to see how it fits together. First, the fact that A is m×p and B is p×n implies that the number of columns of A is equal to the number of rows of B. If so, then the size of the product AB is obtained by a sort of cancellation of the “inside” dimensions:

If the inside dimensions are not equal, then the product AB is not defined.

Example 6

If A is a 3×2 matrix and B is a 2×3 matrix, then AB will be a 3×3 matrix, whereas BA will be a 2×2 matrix. If C is a 3×5 matrix and D is a 5×7 matrix, then CD will be a 3×7 matrix, but DC is undefined.

To emphasize the fact that the ijth element of AB is the product of the ith row of A and the jth column of B, we can write

where a1,a2,,am denote the m row vectors of A and b1,b2,,bn denote the n column vectors of B. More briefly, if

A=[a1a2am]andB=[b1b2bn]

in terms of the rows of A and the columns of B, then

(15)AB=[aibj].

Therefore, as mentioned earlier, the ijth element aibj of AB is given in terms of elements of A and B by

aibj=[ai1ai2aip][b1jb2jbpj]=ai1b1j+ai2b2j++aipbpj.

That is,

(16)aibj=k=1paikbkj.

One can visualize “pouring the ith row of A down the jth column of B” until elements match in pairs, then forming the sum of the products of these pairs, to obtain the element cij of the matrix C=AB.

Suggestion

The key to accuracy and confidence in computing matrix products lies in doing it systematically. Always perform your computations in the same order. First calculate the elements of the first row of AB by multiplying the first row of A by the successive columns of B; second, calculate the elements of the second row of AB by multiplying the second row of A by the successive columns of B; and so forth.

Computing systems often are used for the calculation of products of “large” matrices. If the matrices A and B, with appropriate sizes, have been entered—as illustrated in the 3.2 Application—then the Maple command


with(linalg) :  C := multiply(A,B),

or the Mathematica command


C = A.B,

or the Matlab command


C = A*B

immediately yield the product matrix C=AB.

Matrix Equations

If A=[aij] is an m×n coefficient matrix and x=(x1,x2,,xn) is an n×1 variable (column) matrix, then the product Ax is the m×1 matrix

Ax=[a11a12a1na21a22a2nam1am2amn][x1x2xn]=[a11x1+a12x2++a1nxna21x1+a22x2++a2nxnam1x1+am2x2++amnxn]=(?)[b1b2bm]=b.

We therefore see that

(17)Ax=b

if and only if x=(x1,x2,,xn) is a solution of the linear system in (5). Thus, matrix multiplication enables us to “boil down” a system of m scalar equations in n unknowns to the single matrix equation in (17), which is analogous in notation to the single scalar equation ax=b in a single variable x.

Example 7

The system

3x14x2+x3+7x4=104x15x3+2x4=0x1+9x2+2x36x4=5

of three equations in four unknowns is equivalent to the single matrix equation

[341740521926][x1x2x3x4]=[1005].

Matrix Algebra

The definitions of matrix addition and multiplication can be used to establish the rules of matrix algebra listed in the following theorem.

The only verification that is not entirely routine is that of the associative law of multiplication; see Problem 44 for an outline. Each of the others follows quickly from the corresponding law for the ordinary arithmetic of real numbers. As an illustration, we prove the first distributive law. Suppose that A=[aij] is an m×p matrix and that B=[bij] and C=[cij] are p×n matrices. Then

B+C=[bij+cij],

so by (16) the ijth element of the m×n matrix A(B+C) is

(18)k=1paik(bkj+ckj).

The ijth element of the m×n matrix AB+AC is

(19)k=1paikbkj+k=1paikckj=k=1p(aikbkj+aikckj).

But the distributive law for real numbers, a(b+c)=ab+ac, tells us that corresponding terms of the sums in (18) and (19) are equal. Hence, the ijth terms of the two m×n matrices A(B+C) and AB+AC are equal, and so these matrices are equal: A(B+C)=AB+AC.

If a and b are real numbers, then rules such as

(a+b)C=aC+bC,(ab)C=a(bC),a(BC)=(aB)C

are even easier to verify. What all these rules amount to is this: In matrix manipulations, pairs of parentheses can be inserted or deleted in the same ways as in the ordinary algebra of real numbers.

But not all of the rules of “ordinary” algebra carry over to matrix algebra. In Example 5 we saw that multiplication of matrices is not commutative—in general, ABBA. Other exceptions are associated with zero matrices. A zero matrix is one whose elements are all zero, such as

[0000],[000000],[000000],[00].

We ordinarily denote a zero matrix (whatever its size) by 0. It should be clear that for any matrix A,

0+A=A=A+0,A0=0,and0A=0,

where in each case 0 is a zero matrix of appropriate size. Thus zero matrices appear to play a role in the arithmetic of matrices similar to the role of the real number 0 in ordinary arithmetic.

For real numbers, the following two rules are familiar:

The following example shows that matrices do not obey either of these rules.

Example 8

If

A=[41273115],B=[15312423],andC=[34212313],

then BC, but

AB=[2510185]=AC.(Check this!)

Thus the cancellation law does not generally hold for matrices. If

D=BC=[21120110],

then

AD=[0000]=0,

despite the fact that neither A nor D is a zero matrix. See Problems 31–38 for additional ways in which the algebra of matrices differs significantly from the familiar algebra of real numbers.

Recall that an identity matrix is a square matrix I that has ones on its principal diagonal and zeros elsewhere. Identity matrices play a role in matrix arithmetic which is strongly analogous to that of the real number 1, for which a·1=1·a=a for all values of the real number a. For instance, you can check that

[abcd][1001]=[1001][abcd]=[abcd].

Similarly, if

A=[a11a12a13a21a22a23a31a32a33]andI=[100010001],

then AI=IA=A. For instance, the element in the second row and third column of AI is

(a21)(0)+(a22)(0)+(a23)(1)=a23.

If a is a nonzero real number and b=a1, then ab=ba=1. Given a nonzero square matrix A, the question as to whether there exists an inverse matrix B, one such that AB=BA=I, is more complicated and is investigated in Section 3.5.

3.4 Problems

In Problems 1–4, two matrices A and B and two numbers c and d are given. Compute the matrix cA+dB.

  1. A=[3527],B=[1034],c=3,d=4

  2. A=[203156],B=[231715],c=5,d=3

  3. A=[500731],B=[453274],c=2,d=4

  4. A=[210403527],B=[634521079],c=7,d=5

In Problems 5–12, two matrices A and B are given. Calculate whichever of the matrices AB and BA is defined.

  1. A=[2132],B=[4213]

  2. A=[103324235],B=[743152039]

  3. A=[123],B=[345]

  4. A=[103254],B=[301465]

  5. A=[32],B=[023145]

  6. A=[2143],B=[104325]

  7. A=[35],B=[27561423]

  8. A=[1032],B=[2753910]

In Problems 13–16, three matrices A, B, and C are given. Verify by computation of both sides the associative law A(BC)=(AB)C.

  1. A=[3114],B=[2531],C=[0123]

  2. A=[21],B=[2531],C=[65]

  3. A=[32],B=[112],C=[200314]

  4. A=[200314],B=[1132],C=[10123201]

In Problems 17–22, first write each given homogeneous system in the matrix form Ax=0. Then find the solution in vector form, as in Eq. (9).

  1. x15x3+4x4=0x2+2x37x4=0

  2. x13x2+6x4=0x3+9x4=0

  3. x1+3x4x5=0x22x4+6x5=0x3+x48x5=0

  4. x13x2+7x5=0x32x5=0x410x5=0

  5. x1x3+2x4+7x5=0x2+2x33x4+4x5=0

  6. x1x2+7x4+3x5=0x3x42x5=0

Problems 23 through 26 introduce the idea—developed more fully in the next section—of a multiplicative inverse of a square matrix.

  1. Let

    A=[2132],B=[abcd],

    and

    I=[1001].

    Find B so that AB=I=BA as follows: First equate entries on the two sides of the equation AB=I. Then solve the resulting four equations for a, b, c, and d. Finally verify that BA=I as well.

  2. Repeat Problem 23, but with A replaced by the matrix

    A=[3457].
  3. Repeat Problem 23, but with A replaced by the matrix

    A=[5723].
  4. Use the technique of Problem 23 to show that if

    A=[1224],

    then there does not exist a matrix B such that AB=I. Suggestion: Show that the system of four equations in a, b, c, and d is inconsistent.

  5. A diagonal matrix is a square matrix of the form

    [a10000a20000a30000an],

    in which every element off the main diagonal is zero. Show that the product AB of two n×n diagonal matrices A and B is again a diagonal matrix. State a concise rule for quickly computing AB. Is it clear that AB=BA? Explain.

Problems 28 through 30 develop a method of computing powers of a square matrix.

  1. The positive integral powers of a square matrix A are defined as follows:

    A1=A,A2=AA,A3=AA2,A4=AA3,,An+1=AAn,.

    Suppose that r and s are positive integers. Prove that ArAs=Ar+s and that (Ar)s=Ars (in close analogy with the laws of exponents for real numbers).

  2. If A=[abcd], then show that

    A2=(a+d)A(adbc)I,

    where I denotes the 2×2 identity matrix. Thus every 2×2 matrix A satisfies the equation

    A2(trace A)A+(det A)I=0

    where detA=adbc denotes the determinant of the matrix A, and trace A denotes the sum of its diagonal elements. This result is the 2-dimensional case of the Cayley-Hamilton theorem of Section 6.3.

  3. The formula in Problem 29 can be used to compute A2 without an explicit matrix multiplication. It follows that

    A3=(a+d)A2(adbc)A

    without an explicit matrix multiplication,

    A4=(a+d)A3(adbc)A2,

    and so on. Use this method to compute A2, A3, A4, and A5 given

    A=[2112].

Problems 31–38 illustrate ways in which the algebra of matrices is not analogous to the algebra of real numbers.

    1. Suppose that A and B are the matrices of Example 5. Show that (A+B)(AB)A2B2.

    2. Suppose that A and B are square matrices with the property that AB=BA. Show that (A+B)(AB)=A2B2.

    1. Suppose that A and B are the matrices of Example 5. Show that (A+B)2A2+2AB+B2.

    2. Suppose that A and B are square matrices such that AB=BA. Show that (A+B)2=A2+2AB+B2.

  1. Find four different 2×2 matrices A, with each main diagonal element either +1 or 1, such that A2=I.

  2. Find a 2×2 matrix A with each element +1 or 1 such that A2=0. The formula of Problem 29 may be helpful.

  3. Use the formula of Problem 29 to find a 2×2 matrix A such that A0 and AI but such that A2=A.

  4. Find a 2×2 matrix A with each main diagonal element zero such that A2=I.

  5. Find a 2×2 matrix A with each main diagonal element zero such that A2=I.

  6. This is a continuation of the previous two problems. Find two nonzero 2×2 matrices A and B such that A2+B2=0.

  7. Use matrix multiplication to show that if x1 and x2 are two solutions of the homogeneous system Ax=0 and c1 and c2 are real numbers, then c1x1+c2x2 is also a solution.

    1. Use matrix multiplication to show that if x0 is a solution of the homogeneous system Ax=0 and x1 is a solution of the nonhomogeneous system Ax=b, then x0+x1 is also a solution of the nonhomogeneous system.

    2. Suppose that x1 and x2 are solutions of the nonhomogeneous system of part (a). Show that x1x2 is a solution of the homogeneous system Ax=0.

  8. This is a continuation of Problem 32. Show that if A and B are square matrices such that AB=BA, then

    (A+B)3=A3+3A2B+3AB2+B3

    and

    (A+B)4=A4+4A3B+6A2B2+4AB3+B4.
  9. Let

    A=[120012001]=[100010001]+[020002000]=I+N.
    1. Show that N20 but N3=0.

    2. Use the binomial formulas of Problem 41 to compute

      A2=(I+N)2=I+2N+N2,A3=(I+N)3=I+3N+3N2,

      and

      A4=(I+N)4=I+4N+6N2.
  10. Consider the 3×3 matrix

    A=[211121112].

    First verify by direct computation that A2=3A. Then conclude that An+1=3nA for every positive integer n.

  11. Let A=[ahi], B=[bij], and C=[cjk] be matrices of sizes m×n, n×p, and p×q, respectively. To establish the associative law A(BC)=(AB)C, proceed as follows. By Equation (16) the hjth element of AB is

    i=1nahibij.

    By another application of Equation (16), the hkth element of (AB)C is

    j=1p(i=1nahibij)cjk=i=1nj=1pahibijcjk.

    Show similarly that the double sum on the right is also equal to the hkth element of A(BC). Hence the m×q matrices (AB)C and A(BC) are equal.

3.5 Inverses of Matrices

Recall that the n×n identity matrix is the diagonal matrix

(1)I=[1000010000100001]

having ones on its main diagonal and zeros elsewhere. It is not difficult to deduce directly from the definition of the matrix product that I acts like an identity for matrix multiplication:

(2)AI=AandIB=B

if the sizes of A and B are such that the products AI and IB are defined. It is, nevertheless, instructive to derive the identities in (2) formally from the two basic facts about matrix multiplication that we state below. First, recall that the notation

(3)A=[a1a2a3an]

expresses the m×n matrix A in terms of its column vectors a1,a2,a3,,an.

  1. Fact 1 Ax in terms of columns of A

    If A=[a1a2an] and x=(x1,x2,,xn) is an n-vector, then

    (4)Ax=x1a1+x2a2++xnan.

    The reason is that when each row vector of A is multiplied by the column vector x, its jth element is multiplied by xj.

  2. Fact 2 AB in terms of columns of B

    If A is an m×n matrix and B=[b1b2bp] is an n×p matrix, then

    (5)AB=[Ab1Ab2Abp].

    That is, the j th column of AB is the product of A and the j th column of B. The reason is that the elements of the j th column of AB are obtained by multiplying the individual rows of A by the j th column of B.

Example 1

The third column of the product AB of the matrices

A=[210403]andB=[375426365121]

is

Ab3=[210403][532]=[714].

To prove that AI=A, note first that

(6)I=[e1e2en],

where the jth column vector of I is the jth basic unit vector

(7)ej=[010]j th entry:

If A=[a1a2an], then Fact 1 yields

(8)Aej=0a1++1aj++0an=aj.

Hence Fact 2 gives

AI=A[e1e2en]=[Ae1Ae2Aen]=[a1a2an];

that is, AI=A. The proof that IB=B is similar. (See Problems 41 and 42.)

The Inverse Matrix A-1

If a0, then there is a number b=a1=1/a such that ab=ba=1. Given a nonzero matrix A, we therefore wonder whether there is a matrix B such that AB=BA=I. The following two examples show that the answer to this question depends upon the particular matrix A.

Example 2

If

A=[4937]andB=[7934],

then

AB=[4937][7934]=[1001]=I;

BA=I by a similar computation.

Example 3

Let

A=[1326]andB=[abcd].

If the matrix B had the property that AB=BA=I, then

AB=[1326][abcd]=[a3cb3d2a+6c2b+6d]=[1001].

But upon equating corresponding elements of AB and the 2×2 identity matrix in the last line, we find that

a3c=12a+6c=0andb3d=02b+6d=1.

It is clear that these equations are inconsistent. Thus there can exist no 2×2 matrix B such that AB=I.

Thus the matrix A of Example 2 is invertible, whereas the matrix A of Example 3 is not invertible.

A matrix B such that AB=BA=I is called an inverse matrix of the matrix A. The following theorem says that no matrix can have two different inverse matrices.

The unique inverse of an invertible matrix A is denoted by A1. Thus we say in Example 2 that

IfA=[4937]thenA1=[7934].

In the case of a 2×2 matrix A, it is easy to determine whether or not A is invertible and to find A1 if it exists. In Problems 36 and 37 we ask you to verify the following result.

Equation (9) gives us the following prescription for writing the inverse of an invertible 2×2 matrix:

You might check that this is how B=A1 is obtained from A in Example 2 (in which adbc=1).

Example 4

If

A=[4659],

then adbc=3630=60, so

A1=16[9654]=[3215623].

Arbitrary integral powers of a square matrix A are defined as follows, though in the case of a negative exponent we must assume that A is also invertible. If n is a positive integer, we define

A0=IandA1=A;An+1=AnAfor n1;An=(A1)n.

In Problem 28 of Section 3.4, we asked you to verify the laws of exponents

(10)ArAs=Ar+s,(Ar)s=Ars

in the case of positive integral exponents, and Problem 31 of this section deals with the case of negative integral exponents. In Problem 29 we ask you to establish parts (a) and (b) of the following theorem.

In mathematics it is frequently important to note the surprises. The surprise in Eq. (11) is the reversal of the natural order of the factors in the right-hand side. You should now be able to show that

(ABC)1=C1B1A1.

In general, any product of invertible matrices of the same size is again invertible, and the inverse of a product of invertible matrices is the product in reverse order of their inverses.

Example 5

To solve the system

4x1+6x2=65x1+9x2=18,

we use the inverse of the coefficient matrix

A=[4659]

that we found in Example 4. Then Eq. (13) yields

x=A1b=[4659]1[618]=[3215623][618]=[97].

Thus x1=9, x2=7 is the unique solution.

How to Find A-1

Theorem 2 tells us only how to invert 2×2 matrices. The development of a method for inverting larger matrices involves a special class of matrices, which we define next.

Example 6

We obtain some typical elementary matrices as follows.

[1001](3)R1[3001]=E1[100010001](2)R1+R3[100010201]=E2[100010001]SWAP(R1,R2)[010100001]=E3

The three elementary matrices E1, E2, and E3 correspond to three typical elementary row operations.

Now, suppose that the m×m elementary matrix E corresponds to a certain elementary row operation. It turns out that if we perform this same operation on an arbitrary m×n matrix A, we get the product matrix EA that results upon multiplying A on the left by the matrix E. Thus we can carry out an elementary row operation by means of left multiplication by the corresponding elementary matrix. Problems 3840 illustrate typical cases in the proof of the following theorem.

Elementary row operations are reversible. That is, to every elementary row operation there corresponds an inverse elementary row operation that cancels its effects (see Figure 3.5.1). It follows that every elementary matrix is invertible. To see why, let E be a given elementary matrix and let E1 be the elementary matrix corresponding to the inverse of the row operation that transforms I into E. Then the inverse operation transforms E to I, so Theorem 5 implies that E1E=I. We see similarly that EE1=I. Hence, the elementary matrix E is invertible with E1=E1.

FIGURE 3.5.1.

Inverse elementary row operations.

Elementary Row Operation Inverse Operation
(c)Ri 1cRi
SWAP(Ri,Rj) SWAP(Ri,Rj)
(c)Ri+Rj (c)Ri+Rj

Elementary matrices are not ordinarily used for computational purposes; it is simpler to carry out row operations directly than to multiply by elementary matrices. Instead, their principal role is in the proof of the following theorem, which leads in turn to a practical method for inverting matrices.

The proof of Theorem 6 actually tells us how to find the inverse matrix of A. If we invert each side in Eq. (15) (remembering to reverse the order on the right), we get

(16)A1=EkEk1E2E1I.

Because each left multiplication by an elementary matrix is equivalent to performing the corresponding row operation, we see by comparison of Eqs. (14) and (16) that the same sequence of elementary row operations that transforms A into I also transforms I into A1.

As a practical matter, it generally is more convenient to carry out the two reductions—from A to I and from I to A1—in parallel, as illustrated in our next example.

Example 7

Find the inverse of the 3×3 matrix

A=[432563352].

Solution

We want to reduce A to the 3×3 identity matrix I while simultaneously performing the same sequence of row operations on I to obtain A1. In order to carry out this process efficiently, we adjoin I on the right of A to form the 3×6 matrix

[432100563010352001].

We now apply the following sequence of elementary row operations to this 3×6 matrix (designed to transform its left half into the 3×3 identity matrix).

(1)R3+R1[120101563010352001](1)R3+R2[120101211011352001](2)R1+R2[120101051211352001](3)R1+R3[1201010512110112304](2)R2+R3[120101051211010122]SWAP(R2,R3)[120101010122051211](2)R2+R1[100343010122051211](5)R2+R3[1003430101220017119]

Now that we have reduced the left half of the 3×6 matrix to I, we simply examine its right half to see that the inverse of A is

A1=[3431227119].

Remark

Ordinarily, we do not know in advance whether a given square matrix is invertible or not. To find out, we attempt to carry out the reduction process illustrated in Example 7. If we succeed in reducing A to I, then A is invertible and thereby we find A1. Otherwise—if, somewhere along the way, an all-zero row appears in the left half—we conclude that A is not row equivalent to I, and therefore (by Theorem 6) A is not invertible.

Matrix Equations

In certain applications, one needs to solve a system Ax=b of n equations in n unknowns several times in succession—with the same n×n coefficient matrix A each time, but with different constant vectors b1,b2,,bk on the right. Thus we want to find solution vectors x1,x2,,xk such that

(17)Ax1=b1,Ax2=b2,,Axk=bk.

By Fact 2 at the beginning of this section,

[Ax1Ax2Axk]=A[x1x2xk].

So the k equations in (17) are equivalent to the single matrix equation

(18)AX=B,

where

X=[x1x2xk]andB=[b1b2bk].

If A is invertible and we know A1, we can find the n×k matrix of “unknowns” by multiplying each term in Equation (18) on the left by A1:

(19)X=A1B.

Note that this equation is a generalization of Eq. (13) in Theorem 4. If k=1, it usually is simplest to solve the system by Gaussian elimination, but when several different solutions are sought, it may be simpler to find A1 first and then to apply (19).

Example 8

Find a 3×4 matrix X such that

[432563352]X=[312674155241].

Solution

The coefficient matrix is the matrix A whose inverse we found in Example 7, so Eq. (19) yields

X=A1B=[3431227119][312674155241],

and hence

X=[41314115821133394].

By looking at the third columns of B and X, for instance, we see that the solution of

4x1+3x2+2x3=25x1+6x2+3x3=13x1+5x2+2x3=4

is x1=14, x2=8, x3=39.

Nonsingular Matrices

Theorem 6 tells us that the square matrix A is invertible if and only if it is row equivalent to the identity matrix I, and Theorem 4 in Section 3.3 implies that the latter is true if and only if the system Ax=0 has only the trivial solution x=0. A square matrix having these equivalent properties is sometimes called a nonsingular matrix.

The proof of Theorem 7 is a bit long, but it summarizes most of the basic theory of Chapter 1 and is therefore well worth the effort. Indeed, this theorem is one of the central theorems of elementary linear algebra, and we will need to refer to it repeatedly in subsequent chapters.

3.5 Problems

In Problems 1–8, first apply the formulas in (9) to find A1. Then use A1 (as in Example 5) to solve the system Ax=b.

  1. A=[3243],b=[56]

  2. A=[3725],b=[13]

  3. A=[6756],b=[23]

  4. A=[512717],b=[55]

  5. A=[3254],b=[56]

  6. A=[4736],b=[105]

  7. A=[7957],b=[32]

  8. A=[815510],b=[73]

In Problems 9–22, use the method of Example 7 to find the inverse A1 of each given matrix A.

  1. [5645]

  2. [5746]

  3. [151250271]

  4. [1322833106]

  5. [273132379]

  6. [356243235]

  7. [11514133212]

  8. [133112233]

  9. [130121022]

  10. [122301112]

  11. [143145251]

  12. [201103111]

  13. [0010100001203001]

  14. [4011313101203241]

In Problems 23–28, use the method of Example 8 to find a matrix X such that AX=B.

  1. A=[4354],B=[135125]

  2. A=[7687],B=[204053]

  3. A=[141283274],B=[103022110]

  4. A=[151212172],B=[201030102]

  5. A=[123217227],B=[001101011010]

  6. A=[653532342],B=[210213501105]

  7. Verify parts (a) and (b) of Theorem 3.

Problems 30 through 37 explore the properties of matrix inverses.

  1. Suppose that A, B, and C are invertible matrices of the same size. Show that the product ABC is invertible and that (ABC)1=C1B1A1.

  2. Suppose that A is an invertible matrix and that r and s are negative integers. Verify that ArAs=Ar+s and that (Ar)s=Ars.

  3. Prove that if A is an invertible matrix and AB=AC, then B=C. Thus invertible matrices can be canceled.

  4. Let A be an n×n matrix such that Ax=x for every n-vector x. Show that A=I.

  5. Show that a diagonal matrix is invertible if and only if each diagonal element is nonzero. In this case, state concisely how the inverse matrix is obtained.

  6. Let A be an n×n matrix with either a row or a column consisting only of zeros. Show that A is not invertible.

  7. Show that A=[abcd] is not invertible if adbc=0.

  8. Suppose that adbc0 and A1 is defined as in Equation (9). Verify directly that AA1=A1A=I.

Problems 38 through 40 explore the effect of multiplying by an elementary matrix.

  1. Let E be the elementary matrix E1 of Example 6. If A is a 2×2 matrix, show that EA is the result of multiplying the first row of A by 3.

  2. Let E be the elementary matrix E2 of Example 6 and suppose that A is a 3×3 matrix. Show that EA is the result upon adding twice the first row of A to its third row.

  3. Let E be the elementary matrix E3 of Example 6. Show that EA is the result of interchanging the first two rows of the matrix A.

Problems 41 and 42 complete the proof of Eq. (2).

  1. Show that the ith row of the product AB is AiB, where Ai is the ith row of the matrix A.

  2. Apply the result of Problem 41 to show that if B is an m×n matrix and I is the m×m identity matrix, then IB=B.

  3. Suppose that the matrices A and B are row equivalent. Use Theorem 5 to prove that B=GA, where G is a product of elementary matrices.

  4. Show that every invertible matrix is a product of elementary matrices.

  5. Extract from the proof of Theorem 7 a self-contained proof of the following fact: If A and B are square matrices such that AB=I, then A and B are invertible.

  6. Deduce from the result of Problem 45 that if A and B are square matrices whose product AB is invertible, then A and B are themselves invertible.

3.5 Application Automated Solution of Linear Systems

Linear systems with more than two or three equations are most frequently solved with the aid of calculators or computers. If an n×n linear system is written in the matrix form Ax=b, then we need to calculate first the inverse matrix A1 and then the matrix product x=A1b. Suppose the n×n matrix A and the column vector b have been entered (as illustrated in the 3.2 Application). If A is invertible, then the inverse matrix A1 is calculated by the Maple command with(linalg): inverse(A), the Mathematica command Inverse[A], or the Matlab command inv(A). Consequently, the solution vector x is calculated by the Maple command


with(linalg):   x := multiply(inverse(A),b);

or the Mathematica command


x = Inverse[A].b

or the Matlab command


x = inv(A)*b

Figure 3.5.2 illustrates a similar calculator solution of the linear system

3x12x2+7x3+5x4=5052x1+4x2x3+6x4=4355x1+x2+7x33x4=2864x16x28x3+9x4=445

FIGURE 3.5.2.

TI-89 solution of a linear system Ax=b.

for the solution x1=59, x2=13, x3=17, x4=47. This solution is also given by the Wolfram|Alpha query


A = ((3, −2, 7, 5), (2, 4, −1, 6), (5, 1, 7, −3),
     (4, −6, −8, 9)),
b = (505, 435, 286, 445),
inv(A).b

Remark

Whereas the preceding commands illustrate the handy use of conveniently available inverse matrices to solve linear systems, it might be mentioned that modern computer systems employ direct methods—involving Gaussian elimination and still more sophisticated techniques—that are more efficient and numerically reliable to solve a linear system Ax=b without first calculating the inverse matrix A1.

Use an available calculator or computer system to solve the linear systems in Problems 1–6 of the 3.3 Application. The applied problems below are elementary in character—resembling the “word problems” of high school algebra—but might illustrate the practical advantages of automated solutions.

  1. You are walking down the street minding your own business when you spot a small but heavy leather bag lying on the sidewalk. It turns out to contain U.S. Mint American Eagle gold coins of the following types:

    • One-half ounce gold coins that sell for $285 each,

    • One-quarter ounce gold coins that sell for $150 each, and

    • One-tenth ounce gold coins that sell for $70 each.

    A bank receipt found in the bag certifies that it contains 258 such coins with a total weight of 67 ounces and a total value of exactly $40,145. How many coins of each type are there?

  2. Now you really strike it rich! You find a bag containing one-ounce U.S. American Eagle gold coins valued at $550 each, together with half-ounce and quarter-ounce coins valued as in the preceding problem. If this bag contains a total of 365 coins with a total weight of exactly 11 pounds and a total value of $100,130, how many gold coins of each type are there?

  3. A commercial customer orders 81 gallons of paint that contains equal amounts of red paint, green paint, and blue paint—and, hence, could be prepared by mixing 27 gallons of each. However, the store wishes to prepare this order by mixing three types of paint that are already available in large quantity:

    • a reddish paint that is a mixture of 50% red, 25% green, and 25% blue paint;

    • a greenish paint that is 12.5% red, 75% green, and 12.5% blue paint; and

    • a bluish paint that is 20% red, 20% green, and 60% blue paint.

    How many gallons of each are needed to prepare the customer’s order?

  4. Now the paint store receives a really big order—for 244 gallons of paint that is 1/2 red paint, 1/4 green paint, and 1/4 blue paint. The store has three already-mixed types of paint available in large quantity—the greenish paint and the bluish paint of the preceding problem, plus a reddish paint that is 2/3 red paint, 1/6 green paint, and 1/6 blue paint. How many gallons of each must be mixed in order to fill this order?

  5. A tour busload of 45 people attended two Florida theme parks on successive days. On Day 1 the entrance fee was $15 per adult, $8 per child, $12 per senior citizen and the total charge was $558. On Day 2 the entrance fee was $20 per adult, $12 per child, $17 per senior citizen and the total charge was $771. How many adults, children, and senior citizens were on this tour bus?

  6. For some crazy reason, the lunches bought at the first theme park were totaled separately for the adults, children, and seniors. The adults ordered 34 hot dogs, 15 French fries, and 24 soft drinks for a total bill of $70.85. The children ordered 20 hot dogs, 14 French fries, and 15 soft drinks for a total bill of $46.65. The senior citizens ordered 11 hot dogs, 10 French fries, and 12 soft drinks for a total bill of $30.05. What were the prices of a hot dog, an order of French fries, and a soft drink?

  7. A fast-food restaurant sells four types of sandwiches—hamburgers, cheeseburgers, roast beef, and chicken—and has four cash registers. At the end of each day, each cash register tallies the number of each type of sandwich sold, and the total sandwich receipts for the day. The four cash register operators work at different speeds, and one day’s totals were as follows:

    Hamburgers Cheeseburgers Roast Beef Chicken Receipts
    Register 1 37 44 17 23 $232.99
    Register 2 28 35 13 17 $178.97
    Register 3 32 39 19 21 $215.99
    Register 4 47 51 25 29 $294.38

    What was the price of each of the four types of sandwiches?

  8. The fast-food restaurant of the preceding problem adds a ham sandwich to its menu and, because of increased business, it also adds a fifth cash register and reduces prices. After this expansion, one day’s totals were as follows:

    Hamburgers Cheeseburgers Roast Beef Chicken Ham Receipts
    Register 1 41 49 22 26 19 $292.79
    Register 2 34 39 18 20 16 $236.73
    Register 3 36 43 23 24 18 $270.70
    Register 4 49 52 26 31 24 $340.19
    Register 5 52 55 24 28 25 $341.64

    What were the new prices of the five types of sandwiches?

3.6 Determinants

In Theorem 2 of Section 3.5 we saw that the 2×2 matrix

A=[abcd]

is invertible if and only if adbc0. The number adbc is called the determinant of the 2×2 matrix A. There are several common notations for determinants:

(1)detA=det[abcd]=|abcd|=adbc.

In particular, note the vertical bars that distinguish a determinant from a matrix. Sometimes we write det(A) to emphasize that (1) defines a function which associates a number |A| with each 2×2 matrix A.

Example 1

|3746|=(3)(6)(4)(7)=46

Determinants are often used in elementary algebra to solve a 2×2 system of the form

(2)a11x+a12y=b1a21x+a22y=b2.

It follows from Theorems 2 and 7 in Section 3.5 that this system has a unique solution if and only if its coefficient determinant—the determinant of its coefficient matrix A—is nonzero:

(3)detA=|a11a12a21a22|0.

Then Cramer’s rule says that this unique solution is given by

(4)x=|b1a12b2a22||a11a12a21a22|,y=|a11b1a21b2||a11a12a21a22|.

Thus Cramer’s rule gives each of x and y as a quotient of two determinants, the denominator in each case being the determinant of the coefficient matrix. In the numerator for x the coefficients a11 and a21 of x are replaced with the right-side coefficients b1 and b2, whereas in the numerator for y the coefficients of y are replaced with b1 and b2.

The proof of Cramer’s rule is a routine calculation—to verify that evaluation of the determinant in (4) gives the same expressions for x and y as are obtained by solution of (2) by the method of elimination.

Example 2

To apply Cramer’s rule to solve the system

7x+8y=56x+9y=4

with coefficient determinant

|7869|=15,

we simply substitute in the equations in (4) to get

x=|5849|15=1315,y=|7564|15=215.

Higher-Order Determinants

We define 3×3 determinants in terms of 2×2 determinants, 4×4 determinants in terms of 3×3 determinants, and so on. This type of definition—one dimension at a time, with the definition in each dimension depending on its meaning in lower dimensions—is called an inductive definition.

The determinant detA=|aij| of a 3×3 matrix A=[aij] is defined as follows:

(5)|a11a12a13a21a22a23a31a32a33|=a11|a22a23a32a33|a12|a21a23a31a33|+a13|a21a22a31a32|.

Note the single minus sign on the right-hand side. The three 2×2 determinants in (5) are multiplied by the elements a11, a12, and a13 along the first row of the matrix A. Each of these elements a1j is multiplied by the determinant of the 2×2 submatrix of A that remains after the row and column containing a1j are deleted.

Example 3

|523401312|=(5)|0112|(2)|4132|+(3)|4031|=(5)(1)+(2)(5)(3)(4)=27

The definition of higher-order determinants is simplified by the following notation and terminology.

For example, the minor of a12 in a 3×3 matrix is

The minor of a32 in a 4×4 matrix is

According to Eq. (6), the cofactor Aij is obtained by attaching the sign (1)i+j to the minor Mij. This sign is most easily remembered as the one that appears in the ijth position in checkerboard arrays such as

[+++++]and[++++++++].

Note that a plus sign always appears in the upper left corner and that the signs alternate horizontally and vertically. In the 4×4 case, for instance,

A11=+M11,A12=M12,A13=+M13,A14=M14,A21=M21,A22=+M22,A23=M23,A24=+M24,

and so forth.

With this notation, the definition of 3×3 determinants in (5) can be rewritten as

(7)detA=a11M11a12M12+a13M13=a11A11+a12A12+a13A13.

The last formula is the cofactor expansion of det A along the first row of A. Its natural generalization yields the definition of the determinant of an n×n matrix, under the inductive assumption that (n1)×(n1) determinants have already been defined.

In numerical computations, it frequently is more convenient to work first with minors rather than with cofactors and then attach signs in accord with the checkerboard pattern illustrated previously. Note that determinants have been defined only for square matrices.

Example 4

To evaluate the determinant of

A=[2003010074356224],

we observe that there are only two nonzero terms in the cofactor expansion along the first row. We need not compute the cofactors of zeros, because they will be multiplied by zero in computing the determinant; hence

detA=+(2)|100435224|(3)|010743622|.

Each of the 3×3 determinants on the right-hand side has only a single nonzero term in its cofactor expansion along the first row, so

detA=(2)(1)|3524|+(3)(+1)|7362|=(2)(1210)+(3)(14+18)=92.

Note that, if we could expand along the second row in Example 4, there would be only a single 3×3 determinant to evaluate. It is in fact true that a determinant can be evaluated by expansion along any row or column. The proof of the following theorem is included in Appendix B.

The formulas in (9) and (10) provide 2n different cofactor expansions of an n×n determinant. For n=3, for instance, we have

detA=a11A11+a12A12+a13A13=a21A21+a22A22+a23A23=a31A31+a32A32+a33A33}row expansions=a11A11+a21A21+a31A31=a12A12+a22A22+a32A32=a13A13+a23A23+a33A33.}column expansions

In a specific example, we naturally attempt to choose the expansion that requires the least computational labor.

Example 5

To evaluate the determinant of

A=[760932450],

we expand along the third column, because it has only a single nonzero entry. Thus

detA=(2)|7645|=(2)(3524)=22.

In addition to providing ways of evaluating determinants, the theorem on cofactor expansions is a valuable tool for investigating the general properties of determinants. For instance, it follows immediately from the formulas in (9) and (10) that, if the square matrix A has either an all-zero row or an all-zero column, then detA=0. For example, by expanding along the second row we see immediately that

|173324000806241|=0.

Row and Column Properties

We now list seven properties of determinants that simplify their computation. Each of these properties is readily verified directly in the case of 2×2 determinants. Just as our definition of n×n determinants was inductive, the following discussion of Properties 1–7 of determinants is inductive. That is, we suppose that n3 and that these properties have already been verified for (n1)×(n1) determinants.

Property 1: If the n×n matrix B is obtained from A by multiplying a single row (or a column) of A by the constant k, then detB=kdetA.

For instance, if the ith row of A is multiplied by k, then the elements off the ith row of A are unchanged. Hence for each j=1,2,,n, the ijth cofactors of A and B are equal: Aij=Bij. Therefore, expansion of B along the ith row gives

detB=(kai1)Bi1+(kai2)Bi2++(kain)Bin=k(ai1Ai1+ai2Ai2++ainAin),

and thus detB=kdetA.

Property 1 implies simply that a constant can be factored out of a single row or column of a determinant. Thus we see that

|7151729651210|=(3)|75172365410|

by factoring 3 out of the second column.

Property 2: If the n×n matrix B is obtained from A by interchanging two rows (or two columns), then detB=detA.

To see why this is so, suppose (for instance) that the first row is not one of the two that are interchanged (recall that n3). Then for each j=1,2,,n, the cofactor B1j is obtained by interchanging two rows of the cofactor A1j. Therefore, B1j=A1j by Property 2 for (n1)×(n1) determinants. Because b1j=a1j for each j, it follows by expanding along the first row that

detB=b11B11+b12B12++b1nB1n=a11(A11)+a12(A12)++a1n(A1n)=(a11A11+a12A12++a1nA1n),

and thus detB=detA.

Property 3: If two rows (or two columns) of the n×n matrix A are identical, then detA=0.

To see why, let B denote the matrix obtained by interchanging the two identical rows of A. Then B=A, so detB=detA. But Property 2 implies that detB=detA. Thus detA=detA, and it follows immediately that detA=0.

Property 4: Suppose that the n×n matrices A1, A2, and B are identical except for their ith rows—that is, the other n1 rows of the three matrices are identical—and that the ith row of B is the sum of the ith rows of A1 and A2. Then

detB=detA1+detA2.

This result also holds if columns are involved instead of rows.

Property 4 is readily established by expanding B along its ith row. In Problem 45 we ask you to supply the details for a typical case. The main importance (at this point) of Property 4 is that it implies the following property relating determinants and elementary row operations.

Property 5: If the n×n matrix B is obtained by adding a constant multiple of one row (or column) of A to another row (or column) of A, then detB=detA.

Thus the value of a determinant is not changed either by the type of elementary row operation described or by the corresponding type of elementary column operation. The following computation with 3×3 matrices illustrates the general proof of Property 5. Let

A=[a11a12a13a21a22a23a31a32a33]andB=[a11a12a13+ka11a21a22a23+ka21a31a32a33+ka31].

So B is the result of adding k times the first column of A to its third column. Then

(11)detB=|a11a12a13+ka11a21a22a23+ka21a31a32a33+ka31|=|a11a12a13a21a22a22a31a32a32|+k|a11a12a11a21a22a21a31a32a31|.

Here we first applied Property 4 and then factored k out of the second summand with the aid of Property 1. Now the first determinant on the right-hand side in (11) is simply detA, whereas the second determinant is zero (by Property 3—its first and third columns are identical). We therefore have shown that detB=detA, as desired.

Although we use only elementary row operations in reducing a matrix to echelon form, Properties 1, 2, and 5 imply that we may use both elementary row operations and the analogous elementary column operations in simplifying the evaluation of determinants.

Example 6

The matrix

A=[2341423101]

has no zero elements to simplify the computation of its determinant as it stands, but we notice that we can “knock out” the first two elements of its third column by adding twice the first column to the third column. This gives

detA=|2341423101|=|2301403107|=(+7)|2314|=35.

The moral of the example is this: Evaluate determinants with your eyes open.

An upper triangular matrix is a square matrix having only zeros below its main diagonal. A lower triangular matrix is a square matrix having only zeros above its main diagonal. A triangular matrix is one that is either upper triangular or lower triangular, and thus looks like

[1610058007]or[100370465].

The next property tells us that determinants of triangular matrices are especially easy to evaluate.

Property 6: The determinant of a triangular matrix is equal to the product of its diagonal elements.

The reason is that the determinant of any triangular matrix can be evaluated as in the following computation:

|311920286005170004|=(3)|2860517004|=(3)(2)|51704|=(3)(2)(5)(4)=120.

At each step we expand along the first column and pick up another diagonal element as a factor of the determinant.

Example 7

To evaluate

|2132154210|,

we first add the first row to the second and then subtract twice the first row from the third. This yields

|2132154210|=|213008004|=0.

because we produced a triangular matrix having a zero on its main diagonal. (Can you see an even quicker way to do it by keeping your eyes open?)

The Transpose of a Matrix

The transpose AT of a 2×2 matrix A is obtained by interchanging its off-diagonal elements:

(12)IfA=[abcd]thenAT=[acbd].

More generally, the transpose of the m×n matrix A=[aij] is the n×m matrix AT defined by

(13)AT=[aji].

Note the reversal of subscripts; this means that the element of AT in its jth row and ith column is the element of A in the ith row and the jth column of A. Hence the rows of the transpose AT are (in order) the columns of A, and the columns of AT are the rows of A. Thus we obtain AT by changing the rows of A into columns. For instance,

[205417]T=[240157]

and

[726123504]T=[715220634].

If the matrix A is square, then we get AT by interchanging elements of A that are located symmetrically with respect to its main diagonal. Thus AT is the mirror reflection of A through its main diagonal.

In Problems 47 and 48 we ask you to verify the following properties of the transpose operation (under the assumption that A and B are matrices of appropriate sizes and c is a number):

  1. (AT)T=A;

  2. (A+B)T=AT+BT;

  3. (cA)T=cAT;

  4. (AB)T=BTAT.

Property 7: If A is a square matrix, then det(AT)=detA.

This property of determinants can be verified by writing the cofactor expansion of det A along its first row and the cofactor expansion of det(AT) along its first column. When this is done, we see that the two expansions are identical, because the rows of A are the columns of AT (as in Problem 49).

Determinants and Invertibility

We began this section with the remark that a 2×2 matrix A is invertible if and only if its determinant is nonzero: |A|0. This result holds more generally for n×n matrices.

This theorem (along with the others in this section) is proved in Appendix B. So now we can add the statement that detA0 to the list of equivalent properties of nonsingular matrices stated in Theorem 7 of Section 3.5. Indeed, some texts define the square matrix A to be nonsingular if and only if detA0.

Example 8

According to Problem 61, the determinant of the matrix

V=[1aa21bb21cc2]

is

|V|=(ba)(ca)(cb).

Note that |V|0 if and only if the three numbers a, b, and c are distinct. Hence it follows from Theorem 2 that V is invertible if and only if a, b, and c are distinct. For instance, with a=2, b=3, and c=4 we see that the matrix

[1241391416]

is invertible.

The connection between nonzero determinants and matrix invertibility is closely related to the fact that the determinant function “respects” matrix multiplication in the sense of the following theorem.

The fact that |AB|=|A||B| seems so natural that we might fail to note that it is also quite remarkable. Contrast the simplicity of the equation |AB|=|A||B| with the complexity of the seemingly unrelated definitions of determinants and matrix products. For another contrast, we can mention that |A+B| is generally not equal to |A|+|B|. (Pick a pair of 2×2 matrices and calculate their determinants and that of their sum.)

As a first application of Theorem 3, we can calculate the determinant of the inverse of an invertible matrix A:

AA1=I,

so

|A||A1|=|AA1|=|I|=1,

and therefore

(15)|A1|=1|A|.

Now |A|0 because A is invertible. Thus det(A1) is the reciprocal of det A.

Example 9

The determinant of

A=[2220423104143131]

happens to be |A|=3200. Hence A is invertible, and without finding A1 we know from Eq. (15) that |A1|=1/320.

Cramer’s Rule for n×n Systems

Suppose that we need to solve the n×n linear system

(16)Ax=b,

where

A=[aij],x=[x1x2xn],andb=[b1b2bn].

We assume that the coefficient matrix A is invertible, so we know in advance that a unique solution x of (16) exists. The question is how to write x explicitly in terms of the coefficients aij and the constants bi. In the following discussion, we think of x as a fixed (though as yet unknown) vector.

If we denote by a1,a2,,an the column vectors of the n×n matrix A, then

A=[a1a2an].

The desired formula for the ith unknown xi involves the determinant of the matrix [a1ban] that is obtained upon replacing the ith column ai of A with the constant vector b.

For instance, the unique solution (x1,x2,x3) of the 3×3 system

(18)a11x1+a12x2+a13x3=b1a21x1+a22x2+a23x3=b2a31x1+a32x2+a33x3=b3

with |A|=|aij|0 is given by the formulas

(19)x1=1|A||b1a12a13b2a22a23b3a32a33|,x2=1|A||a11b1a13a21b2a23a31b3a33|,x3=1|A||a11a12b1a21a22b2a31a32b3|.

Example 10

Use Cramer’s rule to solve the system

x1+4x2+5x3=24x1+2x2+5x3=33x1+3x2x3=1.

Solution

We find that

|A|=|145425331|=29;

then the formulas in (19) yield

x1=129|245325131|=3329,x2=129|125435311|=3529,

and

x3=129|142423331|=2329.

Inverses and the Adjoint Matrix

The inverse A1 of the invertible n×n matrix A can be found by solving the matrix equation

(20)AX=I.

If we write the coefficient matrix A=[a1a2an], the unknown matrix X=[x1x2xn], and the identity matrix I=[e1e2en] in terms of their columns, then (20) says that

(21)Axj=ej

for j=1,2,,n. Each of these n equations can then be solved explicitly using Cramer’s rule. In particular, Eq. (17) says that the ith element of the column matrix xj is given by

(22)xij=|a1ai1ejai+1an||A|

for i, j=1,2,,n. When these results are assembled (Appendix B), we get an explicit formula for the inverse matrix.

We see in (23) the transpose of the cofactor matrix [Aij] of the n×n matrix A. This transposed cofactor matrix is called the adjoint matrix of A and is denoted by

(24)adj A=[Aij]T=[Aji].

Example 11

Apply the formula in (23) to find the inverse of the matrix

A=[145425331]

of Example 10, in which we saw that |A|=29.

Solution

First we calculate the cofactors of A, arranging our computations in a natural 3×3 array:

A11=+|2531|=17,A12=|4531|=11,A13=+|4233|=18,A21=|4531|=19,A22=+|1531|=14,A23=|1433|=15,A31=+|4525|=10,A32=|1545|=15,A33=+|1442|=14.

(Note the familiar checkerboard pattern of signs.) Thus the cofactor matrix of A is

[Aij]=[171118191415101514].

Next, we interchange rows and columns to obtain the adjoint matrix

adjA=[Aij]T=[171910111415181514].

Finally, in accord with Eq. (23), we divide by |A|=29 to get the inverse matrix

A1=129[171910111415181514].

Computational Efficiency

The amount of labor required to complete a numerical calculation is measured by the number of arithmetical operations it involves. So let us count just the number of multiplications required to evaluate an n×n determinant using cofactor expansions. If n=5, for instance, then the cofactor expansion along a row or column requires computations of five 4×4 determinants. A cofactor expansion of each of these five 4×4 determinants involves four 3×3 determinants. Each of these four 3×3 determinants leads to three 2×2 determinants, and finally each of these 2×2 determinants requires two multiplications for its evaluation. Hence the total number of multiplications required to evaluate the original 5×5 determinant is

5432=5!=120.

In general, the number of multiplications required to evaluate an n×n determinant completely by cofactor expansions is

n!=n(n1)321(n factorial).

Thus, if we ignore the smaller number of additions involved, then our operations count for an n×n determinant is n!. For instance, the evaluation of a 25×25 determinant by cofactor expansion would require about 25!1.55×1025 operations. If we used a supercomputer capable of a billion operations per second this task would require about (1.55×1016)/(365.25×24×3600)492 million years! This same supercomputer would require about 9.64×1047 years—incomparably longer than the estimated age of the universe (perhaps 20 billion years)—to evaluate a 50×50 determinant by cofactor expansion. Contemporary scientific and engineering applications routinely involve matrices of size 1000×1000 (or larger) whose cofactor expansions would require incomprehensibly long times with any conceivable computer.

However, a typical 2000-era desktop computer, using Matlab and performing “only” about 100 million operations per second, calculates the determinant or inverse of a randomly generated 100×100 matrix almost instantaneously. How is this possible? Obviously not by using cofactor expansions!

The answer is that determinants are much more efficiently calculated by Gaussian elimination—that is, by reduction to triangular form, so that only the product of the remaining diagonal elements need be calculated. Similarly, the inverse of the invertible matrix A is much more efficiently calculated by reduction of the augmented matrix [AI] to reduced echelon form (as in Section 3.5) than by use of Cramer’s rule (as in Theorem 5 here). A careful count reveals that the number of arithmetic operations required to reduce an n×n matrix to echelon form is of the order of n3 rather than n!. As indicated in the table of Fig. 3.6.1, n3 is dramatically smaller than n! if n is fairly large. Consequently, cofactor expansions of determinants and Cramer’s rule for the solution of systems are primarily of theoretical importance, and are seldom used in numerical problems where n is greater than 3 or 4.

FIGURE 3.6.1.

Approximate operations counts for evaluating an n×n determinant by Gaussian elimination (23n3) and by cofactor expansion (n!).

n 5 10 25 50 100
23n3 83 667 10,417 83,333 666,667
n! 120 3,628,800 1.55×1025 3.04×1064 9.33×10157

3.6 Problems

Use cofactor expansions to evaluate the determinants in Problems 1–6. Expand along the row or column that minimizes the amount of computation that is required.

  1. |003400050|

  2. |210121012|

  3. |10002050369840107|

  4. |5118732623000304017|

  5. |0010020000000300000405000|

  6. |3011502413650050076917700820|

In Problems 7–12, evaluate each given determinant after first simplifying the computation (as in Example 6) by adding an appropriate multiple of some row or column to another.

  1. |111222333|

  2. |234231327|

  3. |32505176412|

  4. |3651242512|

  5. |1234056700892469|

  6. |2003011112005134007|

Use the method of elimination to evaluate the determinants in Problems 13–20.

  1. |441122143|

  2. |422315543|

  3. |254531145|

  4. |242541421|

  5. |2331043321130432|

  6. |1441012233140132|

  7. |1003012023230333|

  8. |1211213301231424|

Use Cramer’s rule to solve the systems in Problems 21–32.

  1. 3x+4y=25x+7y=1

  2. 5x+8y=38x+13y=5

  3. 17x+7y=612x+5y=4

  4. 11x+15y=108x+11y=7

  5. 5x+6y=123x+4y=6

  6. 6x+7y=38x+9y=4

  7. 5x1+2x22x3=1x1+5x23x3=25x13x2+5x3=2

  8. 5x1+4x22x3=42x1+3x3=22x1x2+x3=1

  9. 3x1x25x3=34x14x23x3=4x15x3=2

  10. x1+4x2+2x3=34x1+2x2+x3=12x12x25x3=3

  11. 2x15x3=34x15x2+3x3=32x1+x2+x3=1

  12. 3x1+4x23x3=53x12x2+4x3=73x1+2x2x3=3

Apply Theorem 5 to find the inverse A1 of each matrix A given in Problems 33–40.

  1. [522153531]

  2. [203542211]

  3. [352234505]

  4. [443315105]

  5. [415245331]

  6. [343321324]

  7. [323032235]

  8. [243231503]

  9. Show that (AB)T=BTAT if A and B are arbitrary 2×2 matrices.

  10. Consider the 2×2 matrices

    A=[abcd]andB=[xy],

    where x and y denote the row vectors of B. Then the product AB can be written in the form

    AB=[ax+bycx+dy].

    Use this expression and the properties of determinants to show that

    detAB=(adbc)|xy|=(detA)(detB).

    Thus the determinant of a product of 2×2 matrices is equal to the product of their determinants.

Each of Problems 43–46 lists a special case of one of Property 1 through Property 5. Verify it by expanding the determinant on the left-hand side along an appropriate row or column.

  1. |ka11a12a13ka21a22a23ka31a32a33|=k|a11a12a13a21a22a23a31a32a33|

  2. |a21a22a23a11a12a13a31a32a33|=|a11a12a13a21a22a23a31a32a33|

  3. |a1b1c1+d1a2b2c2+d2a3b3c3+d3|=|a1b1c1a2b2c2a3b3c3|+|a1b1d1a2b2d2a3b3d3|

  4. |a11+ka12a12a13a21+ka22a22a23a31+ka32a32a33|=|a11a12a13a21a22a23a31a32a33|

Problems 47 through 49 develop properties of matrix transposes.

  1. Suppose that A and B are matrices of the same size. Show that: (a) (AT)T=A; (b) (cA)T=cAT; and (c) (A+B)T=AT+BT.

  2. Let A and B be matrices such that AB is defined. Show that (AB)T=BTAT. Begin by recalling that the ijth element of AB is obtained by multiplying elements in the ith row of A with those in the jth column of B. What is the ijth element of BTAT?

  3. Let A=[aij] be a 3×3 matrix. Show that det(AT)=detA by expanding det A along its first row and det(AT) along its first column.

  4. Suppose that A2=A. Prove that |A|=0 or |A|=1.

  5. Suppose that An=0 (the zero matrix) for some positive integer n. Prove that |A|=0.

  6. The square matrix A is called orthogonal provided that AT=A1. Show that the determinant of such a matrix must be either +1 or 1.

  7. The matrices A and B are said to be similar provided that A=P1BP for some invertible matrix P. Show that if A and B are similar, then |A|=|B|.

  8. Deduce from Theorems 2 and 3 that if A and B are n×n invertible matrices, then AB is invertible if and only if both A and B are invertible.

  9. Let A and B be n×n matrices. Suppose it is known that either AB=I or BA=I. Use the result of Problem 54 to conclude that B=A1.

  10. Let A be an n×n matrix with detA=1 and with all elements of A integers.

    1. Show that A1 has only integer entries.

    2. Suppose that b is an n-vector with only integer entries. Show that the solution vector x of Ax=b has only integer entries.

  11. Let A be a 3×3 upper triangular matrix with nonzero determinant. Show by explicit computation that A1 is also upper triangular.

  12. Figure 3.6.2 shows an acute triangle with angles A, B, and C and opposite sides a, b, and c. By dropping a perpendicular from each vertex to the opposite side, derive the equations

    ccosB+bcosC=accosA+acosC=bacosB+bcosA=c.

    Regarding these as linear equations in the unknowns cos A, cos B, and cos C, use Cramer’s rule to derive the law of cosines by solving for

    cosA=b2+c2a22bc.

    Thus

    a2=b2+c22bccosA.

    Note that the case A=π/2(90°) reduces to the Pythagorean theorem.

    FIGURE 3.6.2.

    The triangle of Problem 58.

  13. Show that

    |2112|=3and|210121012|=4.
  14. Consider the n×n determinant

    Bn=|210000121000012100000021000012|

    in which each entry on the main diagonal is a 2, each entry on the two adjacent diagonals is a 1, and every other entry is zero.

    1. Expand along the first row to show that

      Bn=2Bn1Bn2.
    2. Prove by induction on n that Bn=n+1 for n2.

Problems 61–64 deal with the Vandermonde determinant

V(x1,x2,,xn)=|1x1x12x1n11x2x22x2n11xnxn2xnn1|

that will play an important role in Section 3.7.

  1. Show by direct computation that V(a,b)=ba and that

    V(a,b,c)=|1aa21bb21cc2|=(ba)(ca)(cb).
  2. The formulas in Problem 61 are the cases n=2 and n=3 of the general formula

    (25)V(x1,x2,,xn)=i,j=1i>jn(xixj).

    The case n=4 is

    V(x1,x2,x3,x4)=(x2x1)(x3x1)(x3x2)×(x4x1)(x4x2)(x4x3).

    Prove this as follows. Given x1, x2, and x3, define the cubic polynomial P(y) to be

    (26)P(y)=|1x1x12x131x2x22x231x3x32x331yy2y3|.

    Because P(x1)=P(x2)=P(x3)=0 (why?), the roots of P(y) are x1, x2, and x3. It follows that

    P(y)=k(yx1)(yx2)(yx3),

    where k is the coefficient of y3 in P(y). Finally, observe that expansion of the 4×4 determinant in (26) along its last row gives k=V(x1,x2,x3) and that V(x1,x2,x3,x4)=P(x4).

  3. Generalize the argument in Problem 62 to prove the formula in (25) by induction on n. Begin with the (n1)st-degree polynomial

    P(y)=|1x1x12x1n11x2x22x2n11xn1xn12xn1n11yy2yn1|.
  4. Use the formula in (25) to evaluate the two determinants given next.

    1. |1111124813927141664|

    2. |11111248124813927|

3.7 Linear Equations and Curve Fitting

Linear Equations and Curve FittingLinear algebra has important applications to the common scientific problem of representing empirical data by means of equations or functions of specified types. We give here only a brief introduction to this extensive subject.

Typically, we begin with a collection of given data points (x0,y0),(x1,y1),,(xn,yn) that are to be represented by a specific type of function y=f(x). For instance, y might be the volume of a sample of gas when its temperature is x. Thus the given data points are the results of experiment or measurement, and we want to determine the curve y=f(x) in the xy-plane so that it passes through each of these points; see Figure 3.7.1. Thus we speak of “fitting” the curve to the data points.

FIGURE 3.7.1.

A curve y=f(x) interpolating (that is, passing through) given data points.

We will confine our attention largely to polynomial curves. A polynomial of degree n is a function of the form

(1)f(x)=a0+a1x+a2x2++anxn,

where the coefficients a0,a1,a2,,an are constants. The data point (xi,yi) lies on the curve y=f(x) provided that f(xi)=yi. The condition that this be so for each i=0,1,2,,n yields the n+1 equations

(2)a0+a1x0+a2(x0)2++an(x0)n=y0a0+a1x1+a2(x1)2++an(x1)n=y1a0+a1x2+a2(x2)2++an(x2)n=y2a0+a1xn+a2(xn)2++an(xn)n=yn.

Because the numbers xi and yi are given, this is a system of n+1 linear equations in the n+1 unknowns a0,a1,a2,,an (the coefficients that determine the polynomial in (1)).

The (n+1)×(n+1) coefficient matrix of the system in (1) is the Vandermonde matrix

(3)A=[1x0(x0)2(x0)n1x1(x1)2(x1)n1x2(x2)2(x2)n1xn(xn)2(xn)n],

whose determinant is discussed in Problems 6163 of Section 3.6. It follows from Eq. (25) there that, if the x-coordinates x0,x1,x2,,xn are distinct, then the matrix A is nonsingular. Hence Theorem 7 in Section 3.5 implies that the system in (2) has a unique solution for the coefficients a0,a1,a2,,an in (1). Thus there is a unique nth degree polynomial that fits the n+1 given data points. We call it an interpolating polynomial, and say that it interpolates the given points.

Example 1

Find a cubic polynomial of the form

y=A+Bx+Cx2+Dx3

that interpolates the data points (1,4), (1,2), (2,1), and (3, 16).

Solution

In a particular problem, it generally is simpler to use distinct capital letters rather than subscripted symbols to denote the coefficients. Here we want to find the values of A, B, C, and D so that y(1)=4, y(1)=2, y(2)=1, and y(3)=16. These conditions yield the four linear equations

AB+CD=4A+B+C+D=2A+2B+4C+8D=1A+3B+9C+27D=16.

We readily reduce this system to the echelon form

AB+CD=4B+D=1C+2D=0D=2,

and then back substitution yields A=7, B=3, C=4, and D=2. Thus the desired cubic polynomial is

y=73x4x2+2x3.

The graph of this cubic is shown in Fig. 3.7.2, along with the four original data points.

FIGURE 3.7.2.

Cubic curve through the four data points of Example 1.

Modeling World Population Growth

As a concrete example of interpolation, we consider the growth of the world’s human population. The table in Fig. 3.7.3 shows the total world population (in billions) at 5-year intervals. Actual populations are shown for the years 1975–2010. The figures listed for the years 2015–2040 are the world populations that were predicted by the United Nations on the basis of detailed demographic analysis of population trends during the 20th century on a country-by-country basis throughout the world. Each entry of the final column of this table gives the average annual percentage growth rate during the preceding 5-year period. For instance, 4.062(1.018)54.44 for the growth during the 5-year period 1975–1980, so the average annual growth during this period is about 1.8%.

FIGURE 3.7.3.

World population data.

Year World Population (billions) Percent Growth
1975 4.062
1980 4.440 1.80%
1985 4.853 1.79%
1990 5.310 1.82%
1995 5.735 1.55%
2000 6.127 1.33%
2005 6.520 1.25%
2010 6.930 1.23%
2015 7.349 1.18%
2020 7.758 1.09%
2025 8.142 0.97%
2030 8.501 0.87%
2035 8.839 0.78%
2040 9.157 0.71%

We see that the world population grew at an annual rate of about 1.8% during the 1980s, but the rate of growth has slowed since then, and it is expected to slow even more during the coming decades of the 21st century. In particular, the growth of the world population at the present time in history is not natural or exponential in character—that characterization would imply a constant percentage rate of growth. We explore the possibility of interpolating world population data with polynomial models that might be usable to predict future populations. It seems natural to expect better results with higher-degree interpolating polynomials. Let’s see whether this is so.

Example 2

First, we fit a linear polynomial P1(t)=a+bt (with t=0 in 1900) to the 1995 and 2005 world population values. We need only solve the equations

a+95b=5.735a+105b=6.520

for a=1.7225, b=0.0785. Thus our linear interpolating polynomial is

(4)P1(t)=1.7225+0.0785t.

Example 3

Now let’s fit a quadratic polynomial P2(t)=a+bt+ct2 (with t=0 in 1900) to the 1995, 2000, and 2005 world population values. With the three data points (95,5.735), (100,6.127), and (105,6.520), the system in (2) yields the equations

a+95b+952c=5.735a+100b+1002c=6.127a+105b+1052c=6.520

having the calculator solution a=1.523, b=0.0745, c=0.00002 (Fig. 3.7.4). Thus our quadratic interpolating polynomial is

(5)P2(t)=1.523+0.0745t+0.00002t2.

FIGURE 3.7.4.

TI-84 Plus CE calculator solution of the 3×3 system in Example 3.

Example 4

Next we fit a cubic polynomial P3(t)=a+bt+ct2+dt3 (with t=0 in 1900) to the 1995, 2000, 2005, and 2010 world population values. With the four data points (95, 5.735), (100, 6.127), (105, 6.520), and (110, 6.930), the system in (2) yields the four equations

5.735=a+95b+952c+953d6.127=a+100b+1002c+1003d6.520=a+105b+1052c+1053d6.930=a+110b+1102c+1103d.

As in the 3.5 Application, a calculator or computer yields the solution

[abcd]=[195952953110010021003110510521053111011021103]1[5.7356.1276.5206.930]=[22.8030.7139670.006380.000021333].

Thus our cubic interpolating polynomial is

(6)P3(t)=22.803+0.713967t0.00638t2+0.000021333t3.

Example 5

In order to fit a fourth-degree population model of the form

P4(t)=a+bt+ct2+dt3+et4

to the 1990-1995-2000-2005-2010 world population data, we need to solve the linear system

[190902903904195952953954110010021003100411051052105310541110110211031104][abcde]=[5.3105.7356.1276.5206.930]

to find the values of the coefficients a, b, c, d, and e. The result is

(7)P4(t)=154.473+5.867667t0.08195t2+0.00051333t30.0000012t4.

The table in Fig. 3.7.5 compares our linear, quadratic, cubic, and quartic predictions with the “correct” United Nations prediction for the year 2030. Each “error” in the third column of this table is the amount by which the corresponding prediction undershoots (positive error) or overshoots (negative error) the U.N. prediction. We see that the quadratic prediction is better than the linear but also markedly better than the cubic prediction. The quartic prediction is an improvement over the cubic, yet still not as good as the quadratic. Thus there is at best an uncertain relationship between the degree of the polynomial model and the accuracy of its predictions.

FIGURE 3.7.5.

Predictions of the 2030 world population.

Year 2030 Prediction Error
Linear 8.482 +0.019
Quadratic 8.500 +0.001
Cubic 9.060 0.559
Quartic 8.430 +0.071
United Nations 8.501

Figure 3.7.6 shows the U.N. world population data points for the years 1975 through 2040, together with the plots of the quadratic, cubic, and quartic population functions of Examples 3, 4, and 5. (The plot of the linear population function of Example 2 is virtually indistinguishable from that of the quadratic function for the values of t shown in the figure.) It looks as though the more work we do to find a polynomial fitting selected data points, the less we get for our effort. It is certainly true in this figure that—outside the interval from 1990 to 2010—the higher the degree of the polynomial, the worse it appears to fit the given data points. The issue here is the difference between

FIGURE 3.7.6.

Comparison of world population data and the interpolating polynomials Pn(t) for n=2,3,4.

All four of our polynomials appear to do a good job of interpolating but, somewhat paradoxically, the higher the degree, the worse the apparent accuracy of extrapolation. The highly questionable accuracy of data extrapolation outside the interval of interpolation has significant implications. For instance, consider a news report that when a certain alleged carcinogen was fed to mice in sufficient amounts to kill an elephant, the mice developed cancer. It is then argued that moderate amounts of this carcinogen may cause cancer in humans; or that if 1 part per billion of this carcinogen in the environment kills 1 person, then 1 part per million (a thousand times as much) will kill 1000 people. Such arguments are common, but they may well be cases of extrapolation beyond the range of accuracy. The bottom line is that interpolation is fairly safe—though hardly fail-safe—but extrapolation is risky.

Geometric Applications

In contrast with population prediction, there are interesting situations where curve fitting is exact. For instance, the fact that “two points determine a line” in the plane means that, when we fit the linear function y=a+bx to a given pair of points, we get precisely the one and only straight line in the plane that passes through these points. Similarly, “three points determine a circle,” meaning that there is one and only one circle in the plane that passes through three given noncollinear points. In order to find this particular circle, we recall that the equation of a circle with center (h, k) and radius r is

(8)(xh)2+(yk)2=r2(Fig3.7.7).

FIGURE 3.7.7.

The circle with center (h, k) and radius r.

Simplification gives

(x22hx+h2)+(y22ky+k2)=r2,

that is,

(9)x2+y2+Ax+By+C=0

(where A=2h, B=2k, and C=h2+k2r2) as the general equation of a circle in the plane.

Example 6

Find the equation of the circle that is determined by the points P(1,5), Q(5,3), and R(6, 4).

Solution

Substitution of the xy-coordinates of each of the three points P, Q, and R into (9) gives the three equations

A+5B+C=265A3B+C=346A+4B+C=52.

Reduction of the corresponding augmented coefficient matrix to reduced row-echelon form (Fig. 3.7.8) yields A=4, B=2, and C=20. Thus the equation of the desired circle is

x2+y24x2y20=0.

FIGURE 3.7.8.

TI-89 calculator solution of the 3×3 system in Example 6.

To find its center and radius, we complete the squares in x and y and get

(x2)2+(y1)2=25.

Thus the circle has center (2, 1) and radius 5 (Fig. 3.7.9).

FIGURE 3.7.9.

The circle of Example 6.

Three appropriate points in the plane also determine a central conic with equation of the form

(10)Ax2+Bxy+Cy2=1.

This is a rotated conic section—an ellipse, parabola, or hyperbola—centered at the origin of the xy-coordinate system. Figure 3.7.10 shows a typical rotated ellipse in the plane.

FIGURE 3.7.10.

A rotated central ellipse.

Example 7

Find the equation of the central conic that passes through the same three points P(1,5), Q(5,3), and R(6, 4) of Example 6.

Solution

Substitution of the xy-coordinates of each of the three points P, Q, and R into (10) gives the linear system of three equations

(11)A5B+25C=125A15B+9C=136A+24B+16C=1

in the three unknowns A, B, and C. Reduction of the corresponding augmented coefficient matrix to reduced row-echelon form (Fig. 3.7.11) yields the values

A=27714212,B=17214212,andC=52314212.

FIGURE 3.7.11.

TI-89 calculator solution of the reduced row-echelon form of the augmented coefficient matrix in (11).

If we substitute these coefficient values in (10) and multiply the result by the common denominator 14212, we get the desired equation

(12)277x2172xy+523y2=14212

of our central conic. The computer plot in Fig. 3.7.12 verifies that this rotated ellipse does indeed pass through all three points P, Q, and R.

FIGURE 3.7.12.

Central ellipse passing through the points P, Q, and R of Example 7.

3.7 Problems

In each of Problems 1–10, n+1 data points are given. Find the nth degree polynomial y=f(x) that fits these points.

  1. (1, 1) and (3, 7)

  2. (1,11) and (2,10)

  3. (0, 3), (1, 1), and (2,5)

  4. (1,1), (1,5), and (2, 16)

  5. (1, 3), (2, 3), and (3, 5)

  6. (1,1), (3,13), and (5, 5)

  7. (1,1), (0,0), (1,1), and (2,4)

  8. (1,3), (0,5), (1,7), and (2, 3)

  9. (2,2), (1,2), (1,10), and (2, 26)

  10. (1,27), (1,13), (2,3), and (3,25)

Three points are given in each of Problems 11–14. Find the equation of the circle determined by these points, as well as its center and radius.

  1. (1,1), (6,6), and (7, 5)

  2. (3,4), (5,10), and (9,12)

  3. (1,0), (0,5), and (5,4)

  4. (0, 0), (10, 0), and (7,7)

In Problems 15–18, find an equation of the central ellipse that passes through the three given points.

  1. (0, 5), (5, 0), and (5, 5)

  2. (0, 5), (5, 0), and (10, 10)

  3. (0, 1), (1, 0), and (10, 10)

  4. (0, 4), (3, 0), and (5, 5)

  5. Find a curve of the form y=A+(B/x) that passes through the points (1, 5) and (2, 4).

  6. Find a curve of the form y=Ax+(B/x)+(C/x2) that passes through the points (1, 2), (2, 20), and (4, 41).

A sphere in space with center (h, k, l) and radius r has equation

(xh)2+(yk)2+(zl)2=r2.

Four given points in space suffice to determine the values of h, k, l, and r. In Problems 21 and 22, find the center and radius of the sphere that passes through the four given points P, Q, R, and S. Hint: Substitute each given triple of coordinates into the sphere equation above to obtain four equations that h, k, l, and r must satisfy. To solve these equations, first subtract the first one from each of the other three. How many unknowns are left in the three equations that result?

  1. P(4,6,15), Q(13,5,7), R(5,14,6), S(5,5,9)

  2. P(11,17,17), Q(29,1,15), R(13,1,33), S(19,13,1)

Population Modeling

Problems 23–34 are intended as calculator or computer problems and are based on the U.S. census data in the table of Fig. 3.7.13, listed by national region in millions for the census years 1950–1990. See www.census.gov/population/censusdata/table-16.pdf for further details.

In Problems 23–26, fit a quadratic function to the 1970, 1980, and 1990 population values for the indicated region.

  1. The Northeast

  2. The Midwest

  3. The South

  4. The West

  1. 27–30. The same as Problems 23–26, except fit a cubic polynomial to the 1960, 1970, 1980, and 1990 population data for the indicated region.

  2. 31–34. The same as Problems 23–26, except fit a quartic polynomial to the 1950, 1960, 1970, 1980, and 1990 population data for the indicated region.

Problems 35 through 40 illustrate the use of determinants in fitting polynomial curves to data points.

FIGURE 3.7.13.

Regional population data (in millions) for Problems 23–34.

1950 1960 1970 1980 1990
Northeast 39.478 44.678 49.061 49.137 50.809
Midwest 44.461 51.619 56.590 58.867 59.669
South 47.197 54.973 62.813 75.367 85.446
West 20.190 28.053 34.838 43.171 52.786
U.S. 151.326 179.323 203.302 226.542 248.710
  1. Explain why the determinant equation

    |yx2x1y1x12x11y2x22x21y3x32x31|=0

    fits a quadratic polynomial of the form y=Ax2+Bx+C to the three given points (x1,y1), (x2,y2), and (x3,y3).

  2. Expand the determinant in Problem 35 to find a parabola that interpolates the points (1, 3), (2, 3), and (3, 7).

  3. Explain why the determinant equation

    |x2+y2xy1x12+y12x1y11x22+y22x2y21x32+y32x3y31|=0

    fits a circle of the form x2+y2+Ax+By+C=0 to the three given points (x1,y1), (x2,y2), and (x3,y3).

  4. Expand the determinant in Problem 37 to find the equation of a circle passing through the three points (3,4), (5,10), and (9,12). Then find its center and radius.

  5. Explain why the determinant equation

    |x2xyy21x12x1y1y121x22x2y2y221x32x3y3y321|=0

    fits a central conic equation of the form Ax2+Bxy+Cy2=1 to the three given points (x1,y1), (x2,y2), and (x3,y3).

  6. Expand the determinant in Problem 39 to find the equation of the ellipse passing through the three points (0, 4), (3, 0), and (5, 5).

4 Vector Spaces

4.1 The Vector Space R3

Here we take a fresh look—from the viewpoint of linear algebra—at the familiar 3-dimensional space of the physical world around us (and of multivariable calculus). This review, combining old and new ideas, will provide an introduction to basic concepts that are explored further in subsequent sections of the chapter.

We define three-dimensional coordinate space R3 to be the set of all ordered triples (a, b, c) of real numbers. In ordinary language, the elements of R3 are called points. The numbers a, b, and c are called the coordinates of the point P(a, b, c) and can be regarded as specifying the location of P in a fixed xyz-coordinate system.

The location of the point P can also be specified by means of the arrow or directed line segment (Fig. 4.1.1) that points from the origin (its initial point) to P (its terminal point). Arrows are often used in physics to represent vector quantities, such as force and velocity, that possess both magnitude and direction. For instance, the velocity vector v of a point moving in space may be represented by an arrow that points in the direction of its motion, with the length of the arrow being equal to the speed of the moving point. If we locate this arrow with its initial point at the origin O, then its direction and length are determined by its terminal point (a, b, c). But the arrow is merely a pictorial object; the mathematical object associated with the vector v is simply the point (a, b, c). For this reason, it is customary to use the words point and vector interchangeably for elements of 3-space R3 and thus to refer either to the point (a, b, c) or to the vector (a, b, c).

FIGURE 4.1.1.

The arrow OP representing the vector v=(a,b,c).

Thus “a vector is a point is a vector,” but vector terminology often aids us in visualizing geometric relationships between different points. The point P(a, b, c) determines the vector v=(a,b,c) in R3, and v is represented geometrically (as in Fig. 4.1.1) by the position vector OP from the origin O(0, 0, 0) to P—or equally well by any parallel translate of this arrow. What is important about an arrow usually is not where it is, but how long it is and which way it points.

As in Section 3.4, we adopt the convention that the vector v with components v1, v2, and v3 may be written interchangeably as either

v=(v1,v2,v3)orv=[v1v2v3]

with the column matrix regarded as just another symbol representing one and the same ordered triple of real numbers. Then the following definitions of addition of vectors and of multiplication of vectors by scalars are consistent with the matrix operations defined in Section 3.4.

Thus we add vectors by adding corresponding components—that is, by componentwise addition (just as we add matrices). For instance, the sum of the vectors u=(4,3,5) and v=(5,2,15) is the vector

u+v=(4,3,5)+(5,2,15)=(45,3+2,5+15)=(1,5,10).

The geometric representation of vectors as arrows often converts an algebraic relation into a picture that is readily understood and remembered. Addition of vectors is defined algebraically by Eq. (1). The geometric interpretation of vector addition is the triangle law of addition illustrated in Fig. 4.1.2 (for the case of 2-dimensional vectors in the plane), where the labeled lengths indicate why this interpretation is valid. An equivalent interpretation is the parallelogram law of addition, illustrated in Fig. 4.1.3.

FIGURE 4.1.2.

The triangle law of vector addition.

FIGURE 4.1.3.

The parallelogram law of vector addition.

Multiplication of a vector by a scalar (a real number) is also defined in a componentwise manner.

The length |v| of the vector v=(a,b,c) is defined to be the distance of the point P(a, b, c) from the origin,

(3)|v|=a2+b2+c2.

The length of cv is |c| times the length of v. For instance, if v=(4,3,12), then

(7)v=(74,73,7(12))=(28,21,84),|v|=(4)2+(3)2+(12)2=169=13,and|7v|=|7||v|=713=91.

The geometric interpretation of scalar multiplication is that cv is a vector of length |c||v|, with the same direction as v if c>0 but the opposite direction if c<0 (Fig. 4.1.4).

FIGURE 4.1.4.

The vector cu may have the same direction as u or the opposite direction.

With vector addition and multiplication by scalars defined as in (1) and (2), R3 is a vector space. That is, these operations satisfy the conditions in (a)–(h) of the following theorem.

Of course, 0=(0,0,0) denotes the zero vector in (c) and (d), and u=(1)u in (d). Each of properties (a)–(h) in Theorem 1 is readily verified in a componentwise manner. For instance, if u=(u1,u2,u3) and v=(v1,v2,v3), then use of the (ordinary) distributive law of real numbers gives

r(u+v)=r(u1+v1,u2+v2,u3+v3)=(r(u1+v1),r(u2+v2),r(u3+v3))=(ru1+rv1,ru2+rv2,ru3+rv3)=(ru1,ru2,ru3)+(rv1,rv2,rv3)=r(u1,u2,u3)+r(v1,v2,v3)=ru+rv,

so we have verified property (e).

The Vector Space R2

The familiar coordinate plane R2 is the set of all ordered pairs (a, b) of real numbers. We may regard R2 as the xy-plane in R3 by identifying (a, b) with the vector (a, b, 0) in R3. Then R2 is simply the set of all 3-dimensional vectors that have third component 0.

Clearly, the sum of any two vectors in R2 is again a vector in R2, as is any scalar multiple of a vector in R2. Indeed, vectors in R2 satisfy all the properties of a vector space enumerated in Theorem 1. Consequently, the plane R2 is a vector space in its own right.

The two vectors u and v are collinear—they lie on the same line through the origin and hence point either in the same direction (Fig. 4.1.5) or in opposite directions—if and only if one is a scalar multiple of the other; that is, either

(4)u=cvorv=cu,

for some scalar c. The scalar c merely adjusts the length and direction of one vector to fit the other. If u and v are nonzero vectors, then c=±|u|/|v| in the first relation, with c being positive if the two vectors point in the same direction, negative otherwise.

FIGURE 4.1.5.

Two linearly dependent vectors u and v.

If one of the relations in (4) holds for some scalar c, then we say that the two vectors are linearly dependent. Note that if u=0 while v0, then u=0v but v is not a scalar multiple of u. (Why?) Thus, if precisely one of the two vectors u and v is the zero vector, then u and v are linearly dependent, but only one of the two relations in (4) holds.

If u and v are linearly dependent vectors with u=cv (for instance), then 1u+(c)v=0. Thus there exist scalars a and b not both zero such that

(5)au+bv=0.

Conversely, suppose that Eq. (5) holds with a and b not both zero. If a0 (for instance) then we can solve for

u=bav=cv

with c=b/a, so it follows that u and v are linearly dependent. Therefore, we have proved the following theorem.

The most interesting pairs of vectors are those that are not linearly dependent. The two vectors u and v are said to be linearly independent provided that they are not linearly dependent. Thus u and v are linearly independent if and only if neither is a scalar multiple of the other. By Theorem 2 this is equivalent to the following statement:

(5)The two vectors u and v are linearlyindependent if and only if the relationau+bv=0implies that a=b=0.

Thus the vectors u and v are linearly independent provided that no nontrivial linear combination of them is equal to the zero vector.

Example 1

If u=(3,2),v=(6,4), and w=(5,7), then u and v are linearly dependent, because v=2u. On the other hand, u and w are linearly independent. Here is an argument to establish this fact: Suppose that there were scalars a and b such that

au+bw=0.

Then

a(3,2)+b(5,7)=0,

and thus we get the simultaneous equations

3a+5b=02a7b=0.

It is now easy to show that a=b=0 is the (unique) solution of this system. This shows that whenever

au+bw=0,

it follows that a=b=0. Therefore, u and w are linearly independent.

Alternatively, we could prove that u and w are linearly independent by showing that neither is a scalar multiple of the other (because 5372).

The most important property of linearly independent pairs of plane vectors is this: If u and v are linearly independent vectors in the plane, then any third vector w in R2 can be expressed as a linear combination of u and v. That is, there exist scalars a and b such that w=au+bv (Fig. 4.1.6). This is a statement that the two linearly independent vectors u and v suffice (in an obvious sense) to “generate” the whole plane R2. This general fact—which Section 4.3 discusses in a broader context—is illustrated computationally by the following example.

FIGURE 4.1.6.

The vector w as a linear combination of the two linearly independent vectors u and v.

Example 2

Express the vector w=(11,4) as a linear combination of the vectors u=(3,2) and v=(2,7).

Solution

We want to find numbers a and b such that au+bv=w; that is,

a=[32]+b[27]=[114].

This vector equation is equivalent to the 2×2 linear system

[3227][ab]=[114],

which (using Gaussian elimination or Cramer’s rule) we readily solve for a=5, b=2. Thus w=5u+2v.

Linear Independence in R3

We have said that the two vectors are linearly dependent provided that they lie on the same line through the origin. For three vectors u=(u1,u2,u3), v=(v1,v2,v3), and w=(w1,w2,w3) in space, the analogous condition is that the three points (u1,u2,u3), (v1,v2,v3), and (w1,w2,w3) lie in the same plane through the origin in R3. Given u, v, and w, how can we determine whether the vectors u, v, and w are coplanar? The key to the answer is the following observation: If r and s are scalars, then the parallelogram law of addition implies that the vectors u, v, and ru+sv are coplanar; specifically, they lie in the plane through the origin that is determined by the parallelogram with vertices 0, ru, sv, and ru+sv. Thus any linear combination of u and v is coplanar with u and v. This is the motivation for our next definition.

Note that each of the three equations in (6) implies that there exist three scalars a, b, and c not all zero such that

(7)au+bv+cw=0.

For if w=ru+sv (for instance), then

ru+sv+(1)w=0,

so we can take a=r, b=s, and c=10. Conversely, suppose that (7) holds with a, b, and c not all zero. If c0 (for instance), then we can solve for

w=acubcv=ru+sv

with r=a/c and s=b/c, so it follows that the three vectors u, v, and w are linearly dependent. Therefore, we have proved the following theorem.

The three vectors u, v, and w are called linearly independent provided that they are not linearly dependent. Thus u, v, and w are linearly independent if and only if neither of them is a linear combination of the other two. As a consequence of Theorem 3, this is equivalent to the following statement:

(7)The  vectors u, v, and w are linearlyindependent if and only if the relationau+bv+cw=0implies that a=b=c=0.

Thus the three vectors u, v, and w are linearly independent provided that no nontrivial linear combination of them is equal to the zero vector.

Given two vectors, we can see at a glance whether either is a scalar multiple of the other. By contrast, it is not evident at a glance whether or not three given vectors in R3 are linearly independent. The following theorem provides one way to resolve this question.

Hence, in order to determine whether or not three given vectors u, v, and w are linearly independent, we can calculate the determinant in (8). In practice, however, it is usually more efficient to set up and solve the linear system in (9). If we obtain only the trivial solution a=b=c=0, then the three given vectors are linearly independent. But if we find a nontrivial solution, we then can express one of the vectors as a linear combination of the other two and thus see how the three vectors are linearly dependent.

Example 3

To determine whether the three vectors u=(1,2,3), v=(3,1,2), and w=(5,5,6) are linearly independent or dependent, we need to solve the system

au+bv+cw=[135215326][abc]=[000].

By Gaussian elimination, we readily reduce this system to the echelon form

[135013000][abc]=[000].

Therefore, we can choose c=1, and it follows that b=3c=3 and a=3b5c=4. Therefore,

4u+(3)v+w=0,

and hence u, v, and w are linearly dependent, with w=4u+3v.

Basis Vectors in R3

Perhaps the most familiar triple of linearly independent vectors in R3 consists of the basic unit vectors

(10)i=(1,0,0),j=(0,1,0),andk=(0,0,1).

When represented by arrows with the initial points at the origin, these three vectors point in the positive directions along the three coordinate axes (Fig. 4.1.7). The expression

v=ai+bj+ck=(a,b,c)

FIGURE 4.1.7.

The basic unit vectors i, j, and k.

shows both that

  • the three vectors i, j, and k are linearly independent (because v=0 immediately implies a=b=c=0), and that

  • any vector in R3 can be expressed as a linear combination of i, j, and k.

A basis for R3 is a triple u, v, w of vectors such that every vector t in R3 can be expressed as a linear combination

(11)t=au+bv+cw

of them. That is, given any vector t in R3, there exist scalars a, b, c such that Eq. (11) holds. Thus the unit vectors i, j, and k constitute a basis for R3. The following theorem says that any three linearly independent vectors constitute a basis for R3.

Example 4

In order to express the vector t=(4,20,23) as a combination of the linearly independent vectors u=(1,3,2), v=(2,8,7), and w=(1,7,9), we need to solve the system

[121387279][abc]=[42023]

that we obtain by substitution in Eq. (12). The echelon form found by Gaussian elimination is

[121012001][abc]=[433],

so c=3, b=42c=2, and a=42bc=5. Thus

t=5u2v+3w.

Subspaces of R3

Up until this point, we have used the words line and plane only in an informal or intuitive way. It is now time for us to say precisely what is meant by a line or plane through the origin in R3. Each is an example of a subspace of R3.

The nonempty subset V of R3 is called a subspace of R3 provided that V itself is a vector space under the operations of vector addition and multiplication of vectors by scalars. Suppose that the nonempty subset V of R3 is closed under these operations—that is, that the sum of any two vectors in V is also in V and that every scalar multiple of a vector in V is also in V. Then the vectors in V automatically satisfy properties (a) through (h) of Theorem 1, because these properties are “inherited” from R3; they hold for all vectors in R3, including those in V. Consequently, we see that a nonempty subset V of R3 is a subspace of R3 if and only if it satisfies the following two conditions:

  1. If u and v are vectors in V, then u+v is also in V (closure under addition).

  2. If u is a vector in V and c is a scalar, then cu is in V (closure under multiplication by scalars).

It is immediate that V=R3 is a subspace: R3 is a subspace of itself. At the opposite extreme, the subset V={0}, containing only the zero vector, is also a subspace of R3, because 0+0=0 and c0=0 for every scalar c. Thus V={0} satisfies conditions (i) and (ii). The subspaces {0} and R3 are sometimes called the trivial subspaces of R3 (because the verification that they are subspaces is quite trivial). All subspaces other than {0} and R3 itself are called proper subspaces of R3.

Now we want to show that the proper subspaces of R3 are what we customarily call lines and planes through the origin. Let V be a subspace of R3 that is neither {0} nor R3 itself. There are two cases to consider, depending on whether or not V contains two linearly independent vectors.

Case 1: Suppose that V does not contain two linearly independent vectors. If u is a fixed nonzero vector in V, then, by condition (ii) above, every scalar multiple cu is also in V. Conversely, if v is any other vector in V, then u and v are linearly dependent, so it follows that v=cu for some scalar c. Thus the subspace V is the set of all scalar multiples of the fixed nonzero vector u and is therefore what we call a line through the origin in R3. (See Fig. 4.1.8.)

FIGURE 4.1.8.

The line L spanned by the vector u.

Case 2: Suppose that V contains two linearly independent vectors u and v. It then follows from conditions (i) and (ii) that V contains every linear combination au+bv of u and v. (See Problem 38.) Conversely, let w be any other vector in V. If u, v, w were linearly independent, then, by Theorem 4, V would be all of R3. Therefore, u, v, w are linearly dependent, so it follows that there exist scalars a and b such that w=au+bv. (See Problem 40.) Thus the subspace V is the set of all linear combinations au+bv of the two linearly independent vectors u and v and is therefore what we call a plane through the origin in R3. (See Fig. 4.1.9.)

FIGURE 4.1.9.

The plane P spanned by the vectors u and v.

Subspaces of the coordinate plane R2 are defined similarly—they are the nonempty subsets of R2 that are closed under addition and multiplication by scalars. In Problem 39 we ask you to show that every proper subspace of R2 is a line through the origin.

Example 5

Let V be the set of all vectors (x, y) in R2 such that y=x. Given u and v in V, we may write u=(u,u) and v=(v,v). Then u+v=(u+v,u+v) and cu=(cu,cu) are in V. It follows that V is a subspace of R2.

Example 6

Let V be the set of all vectors (x, y) in R2 such that x+y=1. Thus V is the straight line that passes through the unit points on the x- and y-axes. Then u=(1,0) and v=(0,1) are in V, but the vector u+v=(1,1) is not. It follows that V is not a subspace of R2.

Example 6 illustrates the fact that lines that do not pass through the origin are not subspaces of R2. Because every subspace must contain the zero vector (per Problem 37), only lines and planes that pass through the origin are subspaces of R3.

4.1 Problems

In Problems 1–4, find |ab|, 2a+b, and 3a4b.

  1. a=(2,5,4), b=(1,2,3)

  2. a=(1,0,2), b=(3,4,5)

  3. a=2i3j+5k, b=5i+3j7k

  4. a=2ij, b=j3k

In Problems 5–8, determine whether the given vectors u and v are linearly dependent or linearly independent.

  1. u=(0,2), v=(0,3)

  2. u=(0,2), v=(3,0)

  3. u=(2,2), v=(2,2)

  4. u=(2,2), v=(2,2)

In Problems 9–14, express w as a linear combination of u and v.

  1. u=(1,2), v=(1,3), w=(1,0)

  2. u=(3,4), v=(2,3), w=(0,1)

  3. u=(5,7), v=(2,3), w=(1,1)

  4. u=(4,1), v=(2,1), w=(2,2)

  5. u=(7,5), v=(3,4), w=(5,2)

  6. u=(5,2), v=(6,4), w=(5,6)

In Problems 15–18, apply Theorem 4 (that is, calculate a determinant) to determine whether the given vectors u, v, and w are linearly dependent or independent.

  1. u=(3,1,2), v=(5,4,6), w=(8,3,4)

  2. u=(5,2,4), v=(2,3,5), w=(4,5,7)

  3. u=(1,1,2), v=(3,0,1), w=(1,2,2)

  4. u=(1,1,0), v=(4,3,1), w=(3,2,4)

In Problems 19–24, use the method of Example 3 to determine whether the given vectors u, v, and w are linearly independent or dependent. If they are linearly dependent, find scalars a, b, and c not all zero such that au+bv+cw=0.

  1. u=(2,0,1), v=(3,1,1), w=(0,2,1)

  2. u=(5,5,4), v=(2,3,1), w=(4,1,5)

  3. u=(1,1,2), v=(2,1,6), w=(3,7,2)

  4. u=(1,1,0), v=(5,1,3), w=(0,1,2)

  5. u=(2,0,3), v=(5,4,2), w=(2,1,1)

  6. u=(1,4,5), v=(4,2,5), w=(3,3,1)

In Problems 25–28, express the vector t as a linear combination of the vectors u, v, and w.

  1. t=(2,7,9), u=(1,2,2), v=(3,0,1), w=(1,1,2)

  2. t=(5,30,21), u=(5,2,2), v=(1,5,3), w=(5,3,4)

  3. t=(0,0,19), u=(1,4,3), v=(1,2,2), w=(4,4,1)

  4. t=(7,7,7), u=(2,5,3), v=(4,1,1), w=(1,1,5)

In Problems 29–32, show that the given set V is closed under addition and under multiplication by scalars and is therefore a subspace of R3.

  1. V is the set of all (x, y, z) such that x=0.

  2. V is the set of all (x, y, z) such that x+y+z=0.

  3. V is the set of all (x, y, z) such that 2x=3y.

  4. V is the set of all (x, y, z) such that z=2x+3y.

In Problems 33–36, show that the given set V is not a subspace of R3.

  1. V is the set of all (x, y, z) such that y=1.

  2. V is the set of all (x, y, z) such that x+y+z=3.

  3. V is the set of all (x, y, z) such that z0.

  4. V is the set of all (x, y, z) such that xyz=1.

  5. Show that every subspace V of R3 contains the zero vector 0.

  6. Suppose that V is a subspace of R3. Show that V is closed under the operation of taking linear combinations of pairs of vectors. That is, show that if u and v are in V and a and b are scalars, then au+bv is in V.

  7. Suppose that V is a proper subspace of R2 and that u is a nonzero vector in V. Show that V is the set of all scalar multiples of u and therefore that V is a line through the origin.

  8. Suppose that u, v, and w are vectors in R3 such that u and v are linearly independent but u, v, and w are linearly dependent. Show that there exist scalars a and b such that w=au+bv.

  9. Let V1 and V2 be subspaces of R3. Their intersection V=V1V2 is the set of all vectors that lie both in V1 and in V2. Show that V is a subspace of R3.

4.2 The Vector Space Rn and Subspaces

In Section 4.1 we defined 3-dimensional space R3 to be the set of all triples (x, y, z) of real numbers. This definition provides a mathematical model of the physical space in which we live, because geometric intuition and experience require that the location of every point be specified uniquely by three coordinates.

In science fiction, the fourth dimension often plays a rather exotic role. But there are common and ordinary situations where it is convenient to use four (or even more) coordinates rather than just two or three. For example, suppose we want to describe the motion of two points P and Q that are moving in the plane R2 under the action of some given physical law. (See Fig. 4.2.1.) In order to tell where P and Q are at a given instant, we need to give two coordinates for P and two coordinates for Q. So let us write P(x1,x2) and Q(x3,x4) to indicate these four coordinates. Then the two points P and Q determine a quadruple or 4-tuple (x1,x2,x3,x4) of real numbers, and any such 4-tuple determines a possible pair of locations of P and Q. In this way the set of all pairs of points P and Q in the plane corresponds to the set of all 4-tuples of real numbers. By analogy with our definition of R3, we may define 4-dimensional space R4 to be the set of all such 4-tuples (x1,x2,x3,x4). Then we can specify a pair of points P and Q in R2 by specifying a single point (x1,x2,x3,x4) in R4, and this viewpoint may simplify our analysis of the motions of the original points P and Q. For instance, it may turn out that their coordinates satisfy some single equation such as

3x14x2+2x35x4=0

that is better understood in terms of a single point in R4 than in terms of the separate points P and Q in R2. Finally, note that in this example the fourth dimension is quite tangible—it refers simply to the second coordinate x4 of the point Q.

FIGURE 4.2.1.

Two points P(x1,x2) and Q(x3,x4) in R2.

But the number 4 is no more special in this context than the numbers 2 and 3. To describe similarly the motion of three points in R3, we would need 9 coordinates rather than 4. For instance, we might be studying the sun-earth-moon system in 3-dimensional space, with (x1,x2,x3) denoting the coordinates of the sun, (x4,x5,x6) the coordinates of the earth, and (x7,x8,x9) the coordinates of the moon at a given instant. Then the single list (x1,x2,x3,x4,x5,x6,x7,x8,x9) of 9 coordinates could be used to specify simultaneously the locations of the sun, earth, and moon. Application of Newton’s second law of motion to each of these three bodies would then lead to a system of differential equations involving the dependent variables xi(t), i=1,2,,9 as functions of time t. (We will discuss this very situation in Section 7.6.)

An efficient description of another situation might require fifty, a hundred, or even a thousand coordinates. For instance, an investor’s stock portfolio might contain 50 different high-tech stocks. If xi denotes the value of the ith stock on a particular day, then the state of the whole portfolio is described by the single list (x1,x2,x3,,x49,x50) of 50 separate numbers. And the evolution of this portfolio with time could then be described by the motion of a single point in 50-dimensional space!

An n-tuple of real numbers is an (ordered) list (x1,x2,x3,,xn) of n real numbers. Thus (1, 3, 4, 2) is a 4-tuple and (0,3,7,5,2,1) is a 6-tuple. A 2-tuple is an ordered pair and a 3-tuple is an ordered triple of real numbers, so the following definition generalizes our earlier definitions of the plane R2 and 3-space R3.

The elements of n-space Rn are called points or vectors, and we ordinarily use boldface letters to denote vectors. The ith entry of the vector x=(x1,x2,x3,,xn) is called its ith coordinate or its ith component. For consistency with matrix operations, we agree that

x=(x1,x2,x3,,xn)=[x1x2xn],

so that the n×1 matrix, or column vector, is simply another notation for the same ordered list of n real numbers.

If n>3, we cannot visualize vectors in Rn in the concrete way that we can “see” vectors in R2 and R3. Nevertheless, we saw in Section 4.1 that the geometric properties of R2 and R3 stem ultimately from algebraic operations with vectors, and these algebraic operations can be defined for vectors in Rn by analogy with their definitions in dimensions 2 and 3.

If u=(u1,u2,,un) and v=(v1,v2,,vn) are vectors in Rn, then their sum is the vector u+v given by

(1)u+v=(u1+v1,u2+v2,,un+vn).

Thus addition of vectors in Rn is defined in a componentwise manner, just as in R2 and R3, and we can visualize u+v as a diagonal vector in a parallelogram determined by u and v, as in Fig. 4.2.2. If c is a scalar—a real number—then the scalar multiple cu is also defined in componentwise fashion:

(2)cu=(cu1,cu2,,cun).

FIGURE 4.2.2.

The parallelogram law for addition of vectors, in Rn just as in R2 or R3.

We can visualize cu as a vector collinear with u, with only its length (and possibly its direction along the same line) altered by multiplication by the scalar c. See Fig. 4.2.3.

In Section 4.1 we saw that, for vectors in R2 and R3, the operations in (1) and (2) satisfy a list of properties that constitute the definition of a vector space; the same is true for vectors in Rn.

Definition of a Vector Space

Let V be a set of elements called vectors, in which the operations of addition of vectors and multiplication of vectors by scalars are defined. That is, given vectors u and v in V and a scalar c, the vectors u+v and cu are also in V (so that V is closed under vector addition and multiplication by scalars). Then, with these operations, V is called a vector space provided that—given any vectors u, v, and w in V and any scalars a and b—the following properties hold true:

(a) u+v=v+u (commutativity)
(b) u+(v+w)=(u+v)+w (associativity)
(c) u+0=0+u=u (zero element)
(d) u+(u)=(u)+u=0 (additive inverse)
(e) a(u+v)=au+av (distributivity)
(f) (a+b)u=au+bu
(g) a(bu)=(ab)u
(h) (1)u=u

FIGURE 4.2.3.

Multiplication of the vector u by the scalar c, in Rn just as in R2 or R3.

In property (c), it is meant that there exists a zero vector 0 in V such that u+0=u. The zero vector in Rn is

0=(0,0,,0).

Similarly, property (d) actually means that, given the vector u in V, there exists a vector u in V such that u+(u)=0. In Rn, we clearly have

u=(u1,u2,,un).

The fact that R=R1 satisfies properties (a)–(h), with the real numbers playing the dual roles of scalars and vectors, means that the real line may be regarded as a vector space. If n>1, then each of properties (a)–(h) may be readily verified for Rn by working with components and applying the corresponding properties of real numbers. For example, to verify the commutativity of vector addition in (a), we begin with Eq. (1) and write

u+v=(u1+v1,u2+v2,,un+vn)=(v1+u1,v2+u2,,vn+un)=v+u.

Thus n-space Rn is a vector space with the operations defined in Eqs. (1) and (2). We will understand throughout that our scalars are the real numbers, though in more advanced treatments of linear algebra other scalars (such as the complex numbers) are sometimes used.

Observe that, in the definition of a vector space, nothing is said about the elements of V (the “vectors”) being n-tuples. They can be any objects for which addition and multiplication by scalars are defined and which satisfy properties (a)–(h). Although many of the vectors you will see in this text actually are n-tuples of real numbers, an example in which they are not may help to clarify the definition of a vector space.

Example 1

Let F be the set of all real-valued functions defined on the real number line R. Then each vector in F is a function f such that the real number f(x) is defined for all x in R. Given f and g in F and a real number c, the functions f+g and cf are defined in the natural way,

(f+g)(x)=f(x)+g(x)

and

(cf)(x)=c(f(x)).

Then each of properties (a)–(h) of a vector space follows readily from the corresponding property of the real numbers. For instance, if a is a scalar, then

[a(f+g)](x)=a[(f+g)(x)]=a[f(x)+g(x)]=af(x)+ag(x)=(af+ag)(x).

Thus a(f+g)=af+ag, so F enjoys property (e).

After verification of the other seven properties, we conclude that F is, indeed, a vector space—one that differs in a fundamental way from each of the vector spaces Rn. We naturally call Rn an n-dimensional vector space, and in Section 4.4 we will define the word dimension in such a way that the dimension of Rn actually is n. But the vector space F of functions turns out not to have dimension n for any integer n; it is an example of an infinite-dimensional vector space. (See Section 4.4.)

Subspaces

Let W={0} be the subset of Rn that contains only the zero vector 0. Then W satisfies properties (a)–(h) of a vector space, because each reduces trivially to the equation 0=0. Thus W is a subset of the vector space V=Rn that is itself a vector space. According to the following definition, a subset of a vector space V that is itself a vector space is called a subspace of V.

In order for the subset W to be a subspace of the vector space V, it first must be closed under the operations of vector addition and multiplication by scalars. Then it must satisfy properties (a)–(h) of a vector space. But W “inherits” all these properties from V, because the vectors in W all are vectors in V, and the vectors in V all satisfy properties (a)–(h). The following “subspace criterion” says that, in order to determine whether the subset W is a vector space, we need only check the two closure conditions.

In Section 4.1 we saw that lines through the origin in R2 form subspaces of R2 and that lines and planes through the origin are subspaces of R3. The subspace W of the following example may be regarded as a higher-dimensional “plane” (or “hyperplane”) through the origin in Rn.

Example 2

Let W be the subset of Rn consisting of all those vectors (x1,x2,,xn) whose coordinates satisfy the single homogeneous linear equation

a1x1+a2x2++anxn=0,

where the given coefficients a1,a2,,an are not all zero. If u=(u1,u2,,un) and v=(v1,v2,,vn) are vectors in W, then

a1(u1+v1)+a2(u2+v2)++an(un+vn)=(a1u1+a2u2++anun)+(a1v1+a2+v2++anvn)=0+0=0,

so u+v=(u1+v1,,un+vn) is also in W. If c is a scalar, then

a1(cu1)+a2(cu2)++an(cun)=c(a1u1+a2u2++anun)=(c)(0)=0,

so cu=(cu1,cu2,,cun) is in W. Thus we have shown that W satisfies conditions (i) and (ii) of Theorem 1 and is therefore a subspace of Rn.

In order to apply Theorem 1 to show that W is a subspace of the vector space V, we must show that W satisfies both conditions in the theorem. But to apply Theorem 1 to show that W is not a subspace of V, we need only show either that condition (i) fails or that condition (ii) fails.

Example 3

Let W be the set of all those vectors (x1,x2,x3,x4) in R4 whose four coordinates are all nonnegative: xi0 for i=1, 2, 3, 4. Then it should be clear that the sum of two vectors in W is also a vector in W, because the sum of two nonnegative numbers is nonnegative. Thus W satisfies condition (i) of Theorem 1. But if we take u=(1,1,1,1) in W and c=1, then we find that the scalar multiple

cu=(1)(1,1,1,1)=(1,1,1,1)

is not in W. Thus W fails to satisfy condition (ii) and therefore is not a subspace of R4.

Example 4

Let W be the set of all those vectors (x1,x2,x3,x4) in R4 such that x1x4=0. Now W satisfies condition (ii) of Theorem 1, because x1x4=0 implies that (cx1)(cx4)=0 for any scalar c. But if we take the vectors u=(1,1,0,0) and v=(0,0,1,1) in W, we see that their sum u+v=(1,1,1,1) is not in W. Thus W does not satisfy condition (i) and therefore is not a subspace of R4.

Example 2 implies that the solution set of a homogeneous linear equation

a1x1+a2x2++anxn=0

in n variables is always a subspace of Rn. Theorem 2 further implies that the same is true of a homogeneous system of linear equations. Recall that any such system of m equations in n unknowns can be written in the form Ax=0, where A is the m×n coefficient matrix and x=(x1,x2,,xn) is regarded as a column vector. The solution set of Ax=0 is then the set of all vectors x in Rn that satisfy this equation—that is, the set of all its solution vectors.

Note that, in order to conclude by Theorem 2 that the solution set of a linear system is a subspace, it is necessary for the system to be homogeneous. Indeed, the solution set of a nonhomogeneous linear system

(4)Ax=b

with b0 is never a subspace. For if u were a solution vector of the system in (4), then

A(2u)=2(Au)=2bb

because b0. Thus the scalar multiple 2u of the solution vector u is not a solution vector. Therefore, Theorem 1 implies that the solution set of (4) is not a subspace.

Because the solution set of a homogeneous linear system is a subspace, we often call it the solution space of the system. The subspaces of Rn are the possible solution spaces of a homogeneous linear system with n unknowns, and this is one of the principal reasons for our interest in subspaces of Rn.

At opposite extremes as subspaces of Rn lie the zero subspace {0} and Rn itself. Every other subspace of Rn, each one that is neither {0} nor Rn, is called a proper subspace of Rn. The proper subspaces of Rn play the same role in Rn that lines and planes through the origin play in R3. In the following two sections we develop the tools that are needed to analyze the structure of a given proper subspace of Rn. In particular, given the homogeneous system Ax=0, we ask how we can describe its solution space in a concise and illuminating way, beyond the mere statement that it is the set of all solution vectors of the system. Example 5 illustrates one possible way of doing this.

Example 5

In Example 4 of Section 3.4 we considered the homogeneous system

(5)x1+3x215x3+7x4=0x1+4x219x3+10x4=02x1+5x226x3+11x4=0.

The reduced echelon form of the coefficient matrix of this system is

[103201430000].

Hence x1 and x2 are the leading variables and x3 and x4 are free variables. Back substitution yields the general solution

x3=s,x4=t,x2=4s3t,x1=3s+2t

in terms of arbitrary parameters s and t. Thus a typical solution vector of the system in (5) has the form

x=[x1x2x3x4]=[3s+2t4s3tst]=s[3410]+t[2301].

It follows that the solution space of the system in (5) can be described as the set of all linear combinations of the form

(6)x=su+tv,

where u=(3,4,1,0) and v=(2,3,0,1). Thus we have found two particular solution vectors u and v of our system that completely determine its solution space [by the formula in (6)].

4.2 Problems

In Problems 1–14, a subset W of some n-space Rn is defined by means of a given condition imposed on the typical vector (x1,x2,,xn). Apply Theorem 1 to determine whether or not W is a subspace of Rn.

  1. W is the set of all vectors in R3 such that x3=0.

  2. W is the set of all vectors in R3 such that x1=5x2.

  3. W is the set of all vectors in R3 such that x2=1.

  4. W is the set of all vectors in R3 such that x1+x2+x3=1.

  5. W is the set of all vectors in R4 such that x1+2x2+3x3+4x4=0.

  6. W is the set of all vectors in R4 such that x1=3x3 and x2=4x4.

  7. W is the set of all vectors in R2 such that |x1|=|x2|.

  8. W is the set of all vectors in R2 such that (x1)2+(x2)2=0.

  9. W is the set of all vectors in R2 such that (x1)2+(x2)2=1.

  10. W is the set of all vectors in R2 such that |x1|+|x2|=1.

  11. W is the set of all vectors in R4 such that x1+x2=x3+x4.

  12. W is the set of all vectors in R4 such that x1x2=x3x4.

  13. W is the set of all vectors in R4 such that x1x2x3x4=0.

  14. W is the set of all those vectors in R4 whose components are all nonzero.

In Problems 15–18, apply the method of Example 5 to find two solution vectors u and v such that the solution space is the set of all linear combinations of the form su+tv.

  1. x14x2+x34x4=0x1+2x2+x3+8x4=0x1+x2+x3+6x4=0

  2. x14x23x37x4=02x1+x2+x3+7x4=0x1+2x2+3x3+11x4=0

  3. x1+3x2+8x3x4=0x13x210x3+5x4=0x1+4x2+11x32x4=0

  4. x1+3x2+2x3+5x4x5=02x1+7x2+4x3+11x4+2x5=02x1+6x2+5x3+12x47x5=0

In Problems 19–22, reduce the given system to echelon form to find a single solution vector u such that the solution space is the set of all scalar multiples of u.

  1. x13x25x36x4=02x1+x2+4x3+4x4=0x1+3x2+7x3+x4=0

  2. x1+5x2+x38x4=02x1+5x25x4=02x1+7x2+x39x4=0

  3. x1+7x2+2x33x4=02x1+7x2+x34x4=03x1+5x2x35x4=0

  4. x1+3x2+3x3+3x4=02x1+7x2+5x3x4=02x1+7x2+4x34x4=0

  5. Show that every subspace W of a vector space V contains the zero vector 0.

  6. Apply the properties of a vector space V to show each of the following.

    1. 0u=0 for every u in V.

    2. c0=0 for every scalar c.

    3. (1)u=u for every u in V.

    Do not assume that the vectors in V are n-tuples of real numbers.

  7. Show that the nonempty subset W of a vector space V is a subspace of V if and only if for every pair of vectors u and v in W and every pair of scalars a and b, au+bv is also in W.

  8. Prove: If u is a (fixed) vector in the vector space V, then the set W of all scalar multiples cu of u is a subspace of V.

  9. Let u and v be (fixed) vectors in the vector space V. Show that the set W of all linear combinations au+bv of u and v is a subspace of V.

  10. Suppose that A is an n×n matrix and that k is a (constant) scalar. Show that the set of all vectors x such that Ax=kx is a subspace of Rn.

  11. Let A be an n×n matrix, b be a nonzero vector, and x0 be a solution vector of the system Ax=b. Show that x is a solution of the nonhomogeneous system Ax=b if and only if y=xx0 is a solution of the homogeneous system Ay=0.

  12. Let U and V be subspaces of the vector space W. Their intersection UV is the set of all vectors that are both in U and in V. Show that UV is a subspace of W. If U and V are two planes through the origin in R3, what is UV?

  13. Let U and V be subspaces of the vector space W. Their sum U+V is the set of all vectors w of the form

    w=u+v,

    where u is in U and v is in V. Show that U+V is a subspace of W. If U and V are lines through the origin in R3, what is U+V?

4.3 Linear Combinations and Independence of Vectors

In Example 5 of Section 4.2 we solved the homogeneous linear system

(1)x1+3x215x3+7x4=0x1+4x219x3+10x4=02x1+5x226x3+11x4=0.

We found that its solution space W consists of all those vectors x in R4 that have the form

(2)x=s(3,4,1,0)+t(2,3,0,1).

We therefore can visualize W as the plane in R4 determined by the vectors v1=(3,4,1,0) and v2=(2,3,0,1). The fact that every solution vector is a combination [as in (2)] of the particular solution vectors v1 and v2 gives us a tangible understanding of the solution space W of the system in (1).

More generally, we know from Theorem 2 in Section 4.2 that the solution set V of any m×n homogeneous linear system Ax=0 is a subspace of Rn. In order to understand such a vector space V better, we would like to find a minimal set of vectors v1,v2,,vk in V such that every vector in V is a sum of scalar multiples of these particular vectors.

The vector w is called a linear combination of the vectors v1,v2,,vk provided that there exist scalars c1,c2,,ck such that

(3)w=c1v1+c2v2++ckvk.

Given a vector w in Rn, the problem of determining whether or not w is a linear combination of the vectors v1,v2,,vk amounts to solving a linear system to see whether we can find scalars c1,c2,,ck so that (3) holds.

Example 1

To determine whether the vector w=(2,6,3) in R3 is a linear combination of the vectors v1=(1,2,1) and v2=(3,5,4), we write the equation c1v1+c2v2=w in matrix form:

c1[121]+c2[354]=[263]

—that is,

c1+3c2=22c15c2=6c1+4c2=3.

The augmented coefficient matrix

[132256143]

can be reduced by elementary row operations to echelon form:

[1320120019].

We see now, from the third row, that our system is inconsistent, so the desired scalars c1 and c2 do not exist. Thus w is not a linear combination of v1 and v2.

Example 2

To express the vector w=(7,7,11) as a linear combination of the vectors v1=(1,2,1), v2=(4,1,2), and v3=(3,1,3), we write the equation c1v1+c2v2+c3v3=w in the form

c1[121]+c2[412]+c3[315]=[7711]

—that is,

c14c23c3=72c1c2+c3=7c1+2c2+3c3=11.

The reduced echelon form of the augmented coefficient matrix of this system is

[101501130000].

Thus c3 is a free variable. With c3=t, back substitution yields c1=5t and c2=3t. For instance, t=1 gives c1=4, c2=2, and c3=1, so

w=4v1+2v2+v3.

But t=2 yields c1=7, c2=5, and c3=2, so w can also be expressed as

w=7v1+5v22v3.

We have found not only that w can be expressed as a linear combination of the vectors v1, v2, v3 but also that this can be done in many different ways (one for each choice of the parameter t).

We began this section with the observation that every solution vector of the linear system in (1) is a linear combination of the vectors v1 and v2 that appear in the right-hand side of Eq. (2). A brief way of saying this is that the vectors v1 and v2 span the solution space. More generally, suppose that v1,v2,,vk are vectors in a vector space V. Then we say that the vectors v1,v2,,vk span the vector space V provided that every vector in V is a linear combination of these k vectors. We may also say that the set S={v1,v2,,vk} of vectors is a spanning set for V.

Example 3

The familiar unit vectors i=(1,0,0), j=(0,1,0), and k=(0,0,1) span R3, because every vector x=(x1,x2,x3) in R3 can be expressed as the linear combination

x=x1i+x2j+x3k

of these three vectors i, j, and k.

If the vectors v1,v2,,vk in the vector space V do not span V, we can ask about the subset of V consisting of all those vectors that are linear combinations of v1,v2,,vk. The following theorem implies that this subset is always a subspace of V.

We say that the subspace W of Theorem 1 is the space spanned by the vectors v1,v2,,vk (or is the span of the set S={v1,v2,,vk} of vectors). We sometimes write

W=span(S)=span{v1,v2,,vk}.

Thus Example 3 implies that R3=span{i,j,k}. The question as to whether a given vector w in Rn lies in the subspace span{v1,v2,,vk} reduces to solving a linear system, as illustrated by Examples 1 and 2.

It is easy to verify that the space W=span{v1,v2,,vk} of Theorem 1 is the smallest subspace of V that contains all the vectors v1,v2,,vk—meaning that every other subspace of V that contains these k vectors must also contain W (Problem 30).

Linear Independence

Henceforth, when we solve a homogeneous system of linear equations, we generally will seek a set v1,v2,,vk of solution vectors that span the solution space W of the system. Perhaps the most concrete way to describe a subspace W of a vector space V is to give explicitly a set v1,v2,,vk of vectors that span W. And this type of representation is most useful and desirable (as well as most aesthetically pleasing) when each vector w in W is expressible in a unique way as a linear combination of v1,v2,,vk. (For instance, each vector in R3 is a unique linear combination of the vectors i, j, and k of Example 3.) But Example 2 demonstrates that a vector w may well be expressed in many different ways as a linear combination of given vectors v1,v2,,vk.

Thus not all spanning sets enjoy the uniqueness property that we desire. Two questions arise:

The following definition provides the key to both answers.

Remark

We can immediately verify that any vector w in the subspace W spanned by the linearly independent vectors v1,v2,,vk is uniquely expressible as a linear combination of these vectors. For

w=i=1kaivi=i=1kbiviimplies thati=1k(aibi)vi=0.

Hence, with ci=aibi for each i=1,,k, the linear independence of v1,v2,,vk implies that c1=c2==ck=0. Thus

i=1kaivi=i=1kbiviimplies thatai=bi for i=1,,k,

so we see that w can be expressed in only one way as a combination of the linearly independent vectors v1,v2,,vk.

Example 4

The standard unit vectors

e1=(1,0,0,,0),e2=(0,1,0,,0),en=(0,0,0,,1)

in Rn are linearly independent. The reason is that the equation

c1e1+c2e2++cnen=0

evidently reduces to

(c1,c2,,cn)=(0,0,,0)

and thus has only the trivial solution c1=c2==cn=0.

Example 5

To determine whether the vectors v1=(1,2,2,1), v2=(2,3,4,1), and v3=(3,8,7,5) in R4 are linearly independent, we write the equation c1v1+c2v2+c3v3=0 as the linear system

c1+2c2+3c3=02c1+3c2+8c3=02c1+4c2+7c3=0c1+c2+5c3=0

and then solve for c1, c2, and c3. The augmented coefficient matrix of this system reduces to the echelon form

[1230012000100000],

so we see that the only solution is c1=c2=c3=0. Thus the vectors v1, v2, and v3 are linearly independent.

Observe that linear independence of the vectors v1,v2,,vk actually is a property of the set S={v1,v2,,vk} whose elements are these vectors. Occasionally the phraseology “the set S={v1,v2,,vk} is linearly independent” is more convenient. For instance, any subset of a linearly independent set S={v1,v2,,vk} is a linearly independent set of vectors (Problem 29).

Now we show that the coefficients in a linear combination of the linearly independent vectors v1,v2,,vk are unique. If both

(5)w=a1v1+a2v2++akvk

and

(6)w=b1v1+b2v2++bkvk,

then

a1v1+a2v2++akvk=b1v1+b2v2++bkvk,

so it follows that

(7)(a1b1)v1+(a2b2)v2++(akbk)vk=0.

Because the vectors v1,v2,,vk are linearly independent, each of the coefficients in (7) must vanish. Therefore, a1=b1,a2=b2,,ak=bk, so we have shown that the linear combinations in (5) and (6) actually are identical. Hence, if a vector w is in the set span{v1,v2,,vk}, then it can be expressed in only one way as a linear combination of these linearly independent vectors.

A set of vectors is called linearly dependent provided it is not linearly independent. Hence the vectors v1,v2,,vk are linearly dependent if and only if there exist scalars c1,c2,,ck not all zero such that

(8)c1v1+c2v2++ckvk=0.

In short, a (finite) set of vectors is linearly dependent provided that some nontrivial linear combination of them equals the zero vector.

Example 6

Let v1=(2,1,3), v2=(5,2,4), v3=(3,8,6), and v4=(2,7,4). Then the equation c1v1+c2v2+c3v3+c4v4=0 is equivalent to the linear system

2c1+5c2+3c3+2c4=0c12c2+8c3+7c4=03c1+4c26c34c4=0

of three equations in four unknowns. Because this homogeneous system has more unknowns than equations, Theorem 3 in Section 3.3 implies that it has a nontrivial solution. Therefore we may conclude—without even solving explicitly for c1, c2, c3, and c4—that the vectors v1, v2, v3, and v4 are linearly dependent. (It happens that

2v1v2+3v34v4=0,

as you can verify easily.)

The argument in Example 6 may be generalized in an obvious way to prove that any set of more than n vectors in Rn is linearly dependent. For if k>n, then Eq. (8) is equivalent to a homogeneous linear system with more unknowns (k) than equations (n), so Theorem 3 in Section 3.3 yields a nontrivial solution.

We now look at the way in which the elements of a linearly dependent set of vectors v1,v2,,vk “depend” on one another. We know that there exist scalars c1,c2,,ck not all zero such that

(9)c1v1+c2v2++ckvk=0.

Suppose that the pth coefficient is nonzero: cp0. Then we can solve Eq. (9) for cpvp and next divide by cp to get

(10)vp=a1v1++ap1vp1+ap+1vp+1++akvk,

where ai=ci/cp for ip. Thus at least one of the linearly dependent vectors is a linear combination of the other k1. Conversely, suppose we are given a set of vectors v1,v2,,vk with one of them dependent on the others as in Eq. (10). Then we can transpose all the terms to the left-hand side to get an equation of the form in (9) with cp=10. This shows that the vectors are linearly dependent. Therefore, we have proved that the vectors v1,v2,,vk are linearly dependent if and only if at least one of them is a linear combination of the others.

For instance (as we saw in Section 4.1), two vectors are linearly dependent if and only if one of them is a scalar multiple of the other, in which case the two vectors are collinear. Three vectors are linearly dependent if and only if one of them is a linear combination of the other two, in which case the three vectors are coplanar.

In Theorem 4 of Section 4.1 we saw that the determinant provides a criterion for deciding whether three vectors in R3 are linearly independent: The vectors v1, v2, v3 in R3 are linearly independent if and only if the determinant of the 3×3 matrix

A=[v1v2v3]

is nonzero. The proof given there in the three-dimensional case generalizes readily to the n-dimensional case. Given n vectors v1,v2,,vn in Rn, we consider the n×n matrix

A=[v1v2vn]

having these vectors as its column vectors. Then, by Theorem 2 in Section 3.6, detA0 if and only if A is invertible, in which case the system Ac=0 has only the trivial solution c1=c2==cn=0, so the vectors v1,v2,,vn must be linearly independent.

We saw earlier that a set of more than n vectors in Rn is always linearly dependent. The following theorem shows us how the determinant provides a criterion in the case of fewer than n vectors in Rn.

Rather than including a complete proof, we will simply illustrate the “if” part of Theorem 3 in the case n=5, k=3. Let v1=(a1,a2,a3,a4,a5), v2=(b1,b2,b3,b4,b5), and v3=(c1,c2,c3,c4,c5) be three vectors in R5 such that the 5×3 matrix

A=[a1b1c1a2b2c2a3b3c3a4b4c4a5b5c5]

has a 3×3 submatrix with nonzero determinant. Suppose, for instance, that

|a1b1c1a3b3c3a5b5c5|0.

Then Theorem 2 implies that the three vectors u1=(a1,a3,a5), u2=(b1,b3,b5), and u3=(c1,c3,c5) in R3 are linearly independent. Now suppose that c1v1+c2v2+c3v3=0. Then by deleting the second and fourth components of each vector in this equation, we find that c1u1+c2u2+c3u3=0. But the fact that u1, u2, u3 are linearly independent implies that c1=c2=c3=0, and it now follows that v1, v2, v3 are linearly independent.

4.3 Problems

In Problems 1–8, determine whether the given vectors v1,v2,,vk are linearly independent or linearly dependent. Do this essentially by inspection—that is, without solving a linear system of equations.

  1. v1=(4,2,6,4), v2=(6,3,9,6)

  2. v1=(3,9,3,6), v2=(2,6,2,4)

  3. v1=(3,4), v2=(6,1), v3=(7,5)

  4. v1=(4,2,2), v2=(5,4,3), v3=(4,6,5), v4=(7,9,3)

  5. v1=(1,0,0), v2=(0,2,0), v3=(0,0,3)

  6. v1=(1,0,0), v2=(1,1,0), v3=(1,1,1)

  7. v1=(2,1,0,0), v2=(3,0,1,0), v3=(4,0,0,1)

  8. v1=(1,0,3,0), v2=(0,2,0,4), v3=(1,2,3,4)

In Problems 9–16, express the indicated vector w as a linear combination of the given vectors v1,v2,,vk if this is possible. If not, show that it is impossible.

  1. w=(1,0,7); v1=(5,3,4), v2=(3,2,5)

  2. w=(3,1,2); v1=(3,1,2), v2=(6,2,3)

  3. w=(1,0,0,1); v1=(7,6,4,5), v2=(3,3,2,3)

  4. w=(4,4,3,3); v1=(7,3,1,9), v2=(2,2,1,3)

  5. w=(5,2,2); v1=(1,5,3), v2=(5,3,4)

  6. w=(2,3,2,3); v1=(1,0,0,3), v2=(0,1,2,0), v3=(0,1,1,1)

  7. w=(4,5,6); v1=(2,1,4), v2=(3,0,1), v3=(1,2,1)

  8. w=(7,7,9,11); v1=(2,0,3,1), v2=(4,1,3,2), v3=(1,3,1,3)

In Problems 17–22, three vectors v1, v2, and v3 are given. If they are linearly independent, show this; otherwise find a nontrivial linear combination of them that is equal to the zero vector.

  1. v1=(1,0,1), v2=(2,3,4), v3=(3,5,2)

  2. v1=(2,0,3), v2=(4,5,6), v3=(2,1,3)

  3. v1=(2,0,3,0), v2=(5,4,2,1), v3=(2,1,1,1)

  4. v1=(1,1,1,1), v2=(2,1,1,1), v3=(3,1,4,1)

  5. v1=(3,0,1,2), v2=(1,1,0,1), v3=(1,2,1,0)

  6. v1=(3,9,0,5), v2=(3,0,9,7), v3=(4,7,5,0)

In Problems 23–26, the vectors {vi} are known to be linearly independent. Apply the definition of linear independence to show that the vectors {ui} are also linearly independent.

  1. u1=v1+v2, u2=v1v2

  2. u1=v1+v2, u2=2v1+3v2

  3. u1=v1, u2=v1+2v2, u3=v1+2v2+3v3

  4. u1=v2+v3, u2=v1+v3, u3=v1+v2

  5. Prove: If the (finite) set S of vectors contains the zero vector, then S is linearly dependent.

  6. Prove: If the set S of vectors is linearly dependent and the (finite) set T contains S, then T is also linearly dependent. You may assume that S={v1,v2,,vk} and that T={v1,v2,,vm} with m>k.

  7. Show that if the (finite) set S of vectors is linearly independent, then any subset T of S is also linearly independent.

  8. Suppose that the subspace U of the vector space V contains the vectors v1,v2,,vk. Show that U contains the subspace spanned by these vectors.

  9. Let S and T be sets of vectors in a vector space such that S is a subset of span(T). Show that span(S) is also a subset of span(T).

  10. Let v1,v2,,vk be linearly independent vectors in the set S of vectors. Prove: If no set of more than k vectors in S is linearly independent, then every vector in S is a linear combination of v1,v2,,vk.

In Problems 33–35, let v1,v2,,vk be vectors in Rn and let

A=[v1v2vk]

be the n×k matrix with these vectors as its column vectors.

  1. Prove: If some k×k submatrix of A is the k×k identity matrix, then v1,v2,,vk are linearly independent.

  2. Suppose that k=n, that the vectors v1,v2,,vk are linearly independent, and that B is a nonsingular n×n matrix. Prove that the column vectors of the matrix AB are linearly independent.

  3. Suppose that k<n, that the vectors v1,v2,,vk are linearly independent, and that B is a nonsingular k×k matrix. Use Theorem 3 to show that the column vectors of AB are linearly independent.

4.4 Bases and Dimension for Vector Spaces

An especially useful way of describing the solution space of a homogeneous linear system is to list explicitly a set S of solution vectors such that every solution vector is a unique linear combination of these particular ones. The following definition specifies the properties of such a set S of “basic” solution vectors, and the concept is equally important for vector spaces other than solution spaces.

In short, a basis for the vector space V is a linearly independent spanning set of vectors in V. Thus, if S={v1,v2,,vn} is a basis for V, then any vector w in V can be expressed as a linear combination

(1)w=c1v1+c2v2++cnvn

of the vectors in S, and we saw in Section 4.3 that the linear independence of S implies that the coefficients c1,c2,,cn in (1) are unique. That is, w cannot be expressed differently as a linear combination of the basis vectors v1,v2,,vn.

Example 1

The standard basis for Rn consists of the unit vectors

e1=(1,0,0,,0),e2=(0,1,0,,0),,en=(0,0,0,,1).

If x=(x1,x2,,xn) is a vector in Rn, then

x=x1e1+x2e2++xnen.

Thus the column vectors

e1=[100],e2=[010],,en=[001]

of the n×n identity matrix span Rn, and we noted in Example 4 of Section 4.3 that these standard unit vectors are linearly independent.

Example 2

Let v1,v2,,vn be n linearly independent vectors in Rn. We saw in Section 4.3 that any set of more than n vectors in Rn is linearly dependent. Hence, given a vector w in Rn, there exist scalars c,c1,c2,,cn not all zero such that

(2)cw+c1v1+c2v2++cnvn=0.

If c were zero, then (2) would imply that the vectors v1,v2,,vn are linearly dependent. Hence c0, so Eq. (2) can be solved for w as a linear combination of v1,v2,,vn. Thus the linearly independent vectors v1,v2,,vn also span Rn and therefore constitute a basis for Rn.

Example 2 shows that any set of n linearly independent vectors in Rn is a basis for Rn. By Theorem 2 in Section 4.3, we can therefore determine whether n given vectors v1,v2,,vn form a basis for Rn by calculating the determinant of the n×n matrix

A=[v1v2vn]

with these vectors as its column vectors. They constitute a basis for Rn if and only if detA0.

Example 3

Let v1=(1,1,2,3), v2=(1,1,2,3), v3=(1,1,3,2), and v4=(0,3,1,2). Then we find that

|1110111322313322|=300,

so it follows that {v1,v2,v3,v4} is a basis for R4.

Theorem 1 has the following import: Just as in Rn, a basis for any vector space V contains the largest possible number of linearly independent vectors in V.

Now let S={v1,v2,,vn} and T={w1,w2,,wm} be two different bases for the same vector space V. Because S is a basis and T is linearly independent, Theorem 1 implies that mn. Next, reverse the roles of S and T; the fact that T is a basis and S is linearly independent implies similarly that nm. Hence m=n, so we have proved the following theorem for any vector space with a finite basis.

A nonzero vector space V is called finite dimensional provided that there exists a basis for V consisting of a finite number of vectors from V. In this case the number n of vectors in each basis for V is called the dimension of V, denoted by n=dimV. Then V is an n-dimensional vector space. The standard basis of Example 1 shows that Rn is, indeed, an n-dimensional vector space.

Note that the zero vector space {0} has no basis because it contains no linearly independent set of vectors. (Sometimes it is convenient to adopt the convention that the null set is a basis for {0}.) Here we define dim{0} to be zero. A nonzero vector space that has no finite basis is called infinite dimensional. Infinite dimensional vector spaces are discussed in Section 4.5, but we include an illustrative example of one here.

Example 4

Let P be the set of all polynomials of the form

p(x)=a0+a1x+a2x2++anxn,

where the largest exponent n0 that appears is the degree of the polynomial p(x), and the coefficients a0,a1,a2,,an are real numbers. We add polynomials in P and multiply them by scalars in the usual way—that is, by collecting coefficients of like powers of x. For instance, if

p(x)=3+2x+5x3andq(x)=7+4x+3x2+9x4,

then

(p+q)(x)=(3+7)+(2+4)x+(0+3)x2+(5+0)x3+(0+9)x4=10+6x+3x2+5x3+9x4

and

(7p)(x)=7(3+2x+5x3)=21+14x+35x3.

It is readily verified that, with these operations, P is a vector space. But P has no finite basis. For if p1,p2,,pn are elements of P, then the degree of any linear combination of them is at most the maximum of their degrees. Hence no polynomial in P of higher degree lies in span{p1,p2,,pn}. Thus no finite subset of P spans P, and therefore P is an infinite dimensional vector space.

Here, our concern is with finite-dimensional vector spaces, and we note first that any proper subspace W of a finite-dimensional vector space V is itself finite dimensional, with dimW<dimV. For if dimV=n, let kn be the largest integer such that W contains k linearly independent vectors v1,v2,,vk. Then, by the same argument as in Example 2, we see that {v1,v2,,vk} is a basis for W, so W is finite dimensional with dimW=k, and k<n because W is a proper subspace.

Moreover, an n-dimensional vector space V contains proper subspaces of each dimension k=1,2,,n1. For instance, if {v1,v2,,vn} is a basis for V and k<n, then W=span{v1,v2,,vk} is a subspace of dimension k. Thus R4 contains proper subspaces of dimensions 1, 2, and 3; R5 contains proper subspaces of dimensions 1, 2, 3, and 4; and so on. The proper subspaces of Rn of dimensions 1,2,,n1 are the higher-dimensional analogues of lines and planes through the origin in R3.

Suppose that V is an n-dimensional vector space and that S={v1,v2,,vn} is a set of n vectors in V. Then, in order to show that S is a basis for V, it is not necessary to prove both that S is linearly independent and that S spans V, because it turns out (for n vectors in an n-dimensional vector space) that either property of S implies the other. For instance, if S={v1,v2,,vn} is a set of n linearly independent vectors in V, then Theorem 1 and the argument of Example 2 (once again) imply that S spans V and, hence, is a basis for V. This proves part (a) of Theorem 3. The remaining parts are left to the problems. (See Problems 2732.)

Part (c) of Theorem 3 is often applied in the following form: If W is a k-dimensional subspace of the n-dimensional vector space V, then any basis {v1,v2,,vk} for W can be “extended” to a basis {v1,v2,,vn} for V, consisting of the original basis vectors for W together with nk additional vectors vk+1,vk+2,,vn.

Bases for Solution Spaces

We consider now the homogeneous linear system

(7)Ax=0,

in which A is an m×n matrix, so the system consists of m equations in the n variables x1,x2,,xn. Its solution space W is then a subspace of Rn. We want to determine the dimension of W and, moreover, to find an explicit basis for W. Thus we seek a maximal set of linearly independent solution vectors of (7).

Recall the Gaussian elimination method of Section 3.2. We use elementary row operations to reduce the coefficient matrix A to an echelon matrix E and note the (nonzero) leading entries in the rows of E. The leading variables are those that correspond to the columns of E containing the leading entries. The remaining variables (if any) are the free variables. If there are no free variables, then our system has only the trivial solution, so W={0}. If there are free variables, we set each of them (separately) equal to a parameter and solve (by back substitution) for the leading variables as linear combinations of these parameters. The solution space W is then the set of all solution vectors obtained in this manner (for all possible values of the parameters).

To illustrate the general situation, let us suppose that the leading variables are the first r variables x1,x2,,xr, so the k=nr variables xr+1,xr+2,,xn are free variables. The reduced system Ex=0 then takes the form

(8)b11x1+b12x2++b1rxr++b1nxn=0b22x2++b2rxr++b2nxn=0brrxr++brnxn=00=00=0

with the last mr equations being “trivial.” We set

(9)xr+1=t1,xr+2=t2,,xn=tk

and then solve (by back substitution) the equations in (8) for the leading variables

(10)x1=c11t1+c12t2++c1ktkx2=c21t1+c22t2++c2ktkxr=cr1t1+cr2t2++crktk.

The typical solution vector (x1,x2,,xn) is given in terms of the k parameters t1,t2,,tk by the equations in (9) and (10).

We now choose k particular solution vectors v1,v2,,vk as follows: To get vj we set the jth parameter tj equal to 1 and set all other parameters equal to zero. Then

(11)vj=(c1j,c2j,,crj,0,,1,,0),

with the 1 appearing as the (r+j)th entry. The vectors v1,v2,,vk are the column vectors of the n×k matrix

(12)[c11c12c1kc21c22c2kcr1cr2crk100010001].

Because of the presence of the lower k×k identity matrix, it is clear that the vectors v1,v2,,vk are linearly independent. (See Problem 36.) But Eqs. (9) and (10) show that the typical solution vector x is a linear combination

(13)x=t1v1+t2v2++tkvk.

Therefore, the vectors v1,v2,,vk defined in (11) form a basis for the solution space W of the original system in (7).

The following algorithm summarizes the steps in this procedure.

Example 5

Find a basis for the solution space of the homogeneous linear system

(14)3x1+6x2x35x4+5x5=02x1+4x2x33x4+2x5=03x1+6x22x34x4+x5=0.

Solution

We readily reduce the coefficient matrix A to the echelon form

E=[120230011400000].

The leading entries are in the first and third columns, so the leading variables are x1 and x3; the free variables are x2, x4, and x5. To avoid subscripts, we use r, s, and t rather than t1, t2, and t3 to denote the three parameters. Thus we set

(15)x2=r,x4=s,andx5=t.

Then back substitution in the reduced system

x1+2x22x4+3x5=0x3x4+4x5=0

yields

(16)x1=2r+2s3tandx3=s4t.

The equations in (15) and (16) give the typical solution vector (x1,x2,x3,x4,x5) in terms of the parameters r, s, and t.

With r=1 and s=t=0,

we obtainv1=(2,1,0,0,0).

With s=1 and r=t=0,

we obtainv2=(2,0,1,1,0).

With t=1 and r=s=0,

we obtainv3=(3,0,4,0,1).

Thus the solution space of the system in (14) is a 3-dimensional subspace of R5 with basis {v1,v2,v3}.

4.4 Problems

In Problems 1–8, determine whether or not the given vectors in Rn form a basis for Rn.

  1. v1=(4,7), v2=(5,6)

  2. v1=(3,1,2), v2=(6,2,4), v3=(5,3,1)

  3. v1=(1,7,3), v2=(2,1,4), v3=(6,5,1), v4=(0,7,13)

  4. v1=(3,7,5,2), v2=(1,1,3,4), v3=(7,11,3,13)

  5. v1=(0,7,3), v2=(0,5,4), v3=(0,5,10)

  6. v1=(0,0,1), v2=(0,1,2), v3=(1,2,3)

  7. v1=(0,0,1), v2=(7,4,11), v3=(5,3,13)

  8. v1=(2,0,0,0), v2=(0,3,0,0), v3=(0,0,7,6), v4=(0,0,4,5)

In Problems 9–11, find a basis for the indicated subspace of R3.

  1. The plane with equation x2y+5z=0.

  2. The plane with equation y=z.

  3. The line of intersection of the planes described in Problems 9 and 10.

In Problems 12–14, find a basis for the indicated subspace of R4.

  1. The set of all vectors of the form (a, b, c, d) for which a=b+c+d.

  2. The set of all vectors of the form (a, b, c, d) such that a=3c and b=4d.

  3. The set of all vectors of the form (a, b, c, d) for which a+2b=c+3d=0.

In Problems 15–26, find a basis for the solution space of the given homogeneous linear system.

  1. x12x2+3x3=02x13x2x3=0

  2. x1+3x2+4x3=03x1+8x2+7x3=0

  3. x13x2+2x34x4=02x15x2+7x33x4=0

  4. x1+3x2+4x3+5x4=02x1+6x2+9x3+5x4=0

  5. x13x29x35x4=02x1+x24x3+11x4=0x1+3x2+3x3+13x4=0

  6. x13x210x3+5x4=0x1+4x2+11x32x4=0x1+3x2+8x3x4=0

  7. x14x23x37x4=02x1x2+x3+7x4=0x1+2x2+3x3+11x4=0

  8. x12x23x316x4=02x14x2+x3+17x4=0x12x2+3x3+26x4=0

  9. x1+5x2+13x3+14x4=02x1+5x2+11x3+12x4=02x1+7x2+17x3+19x4=0

  10. x1+3x24x38x4+6x5=0x1+2x3+x4+3x5=02x1+7x210x319x4+13x5=0

  11. x1+2x2+7x39x4+31x5=02x1+4x2+7x311x4+34x5=03x1+6x2+5x311x4+29x5=0

  12. 3x1+x23x3+11x4+10x5=05x1+8x2+2x32x4+7x5=02x1+5x2x4+14x5=0

Problems 27 through 36 further explore independent sets, spanning sets, and bases.

  1. Suppose that S is a set of n linearly independent vectors in the n-dimensional vector space V. Prove that S is a basis for V.

  2. Suppose that S is a set of n vectors that span the n-dimensional vector space V. Prove that S is a basis for V.

  3. Let {v1,v2,,vk} be a basis for the proper subspace W of the vector space V, and suppose that the vector v of V is not in W. Show that the vectors v1,v2,,vk,v are linearly independent.

  4. Use the result of Problem 29 to prove that every linearly independent set of vectors in a finite-dimensional vector space V is contained in a basis for V.

  5. Suppose that the vectors v1,v2,,vk,vk+1 span the vector space V and that vk+1 is a linear combination of v1,v2,,vk. Show that the vectors v1,v2,,vk span V.

  6. Use the result of Problem 31 to prove that every spanning set for a finite-dimensional vector space V contains a basis for V.

  7. Let S be a linearly independent set of vectors in the finite-dimensional vector space V. Then S is called a maximal linearly independent set provided that if any other vector is adjoined to S, then the resulting set is linearly dependent. Prove that every maximal linearly independent set in V is a basis for V.

  8. Let S be a finite set of vectors that span the vector space V. Then S is called a minimal spanning set provided that no proper subset of S spans V. Prove that every minimal spanning set in V is a basis for V.

  9. Let S be a finite set of vectors that span the vector space V. Then S is called a uniquely spanning set provided that each vector in V can be expressed in one and only one way as a linear combination of the vectors in S. Prove that every uniquely spanning set in V is a basis for V.

  10. Apply the definition of linear independence to show directly that the column vectors of the matrix in (12) are linearly independent.

4.5 Row and Column Spaces

In numerous examples we have observed the phenomenon of “disappearing equations” that sometimes occurs when we solve a linear system using the method of Gaussian elimination. The appearance in this process of a trivial equation 0=0 means that one of the original equations was redundant. For instance, in the system

x2y+2z=0x+4y+3z=02x+2y+5z=0,

the third equation provides no additional information about a solution (x, y, z) because it is merely the sum of the first two equations.

Given a homogeneous linear system, it is natural to ask how many of the equations are “irredundant,” and which ones they are. We will see that an answer to this question leads to a natural and simple relation between the number of irredundant equations, the number of unknowns, and the number of linearly independent solutions.

Row Space and Row Rank

The individual equations of the homogeneous linear system Ax=0 correspond to the “row matrices”

[a11a12a1n][a21a22a2n][am1am2amn]

of the m×n matrix A=[aij]. The row vectors of A are the m vectors

(1)r1=(a11,a12,,a1n)r2=(a21,a22,,a2n)rm=(am1,am2,,amn)

in Rn. Recalling from Section 3.4 the convection that n-tuples denote column-vector elements of Rn, we see that the row vectors of A are the transposes of its row matrices; that is,

ri=[ai1ai2ain]=[ai1ai2ain]T

for i=1,2,,m, and thus actually are column vectors (despite being called “row vectors”). The subspace of Rn spanned by the m row vectors r1,r2,,rm is called the row space Row(A) of the matrix A. The dimension of the row space Row(A) is called the row rank of the matrix A.

Example 1

Consider the 4×5 echelon matrix

A=[13253001420001700000].

Its row vectors are r1=(1,3,2,5,3),r2=(0,0,1,4,2),r3=(0,0,0,1,7), and r4=(0,0,0,0,0). We want to show that the row vectors that are not zero are linearly independent. To this end we calculate the linear combination

c1r1+c2r2+c3r3=(c1,3c1,2c1+c2,5c14c2+c3,3c1+2c2+7c3).

If c1r1+c2r2+c3r3=0, then the first component gives c1=0, next the third component 2c1+c2=0 yields c2=0, and finally the fourth component 5c14c2+c3=0 yields c3=0. Thus the vectors r1, r2, and r3 are linearly independent and therefore form a basis for the row space Row(A). Hence Row(A) is a 3-dimensional subspace of R5, and the row rank of A is 3.

It should be apparent that the method used in Example 1 applies to any echelon matrix E=[eij]. If the first column of E isnot all zero, then its nonzero row vectors r1, r2, , rk have the forms

r1=(e11,,e1p,,e1q,),r2=(0,,0,e2p,,e2q,),r3=(0,,0,,e3q,),

and so forth (where e11, e2p, and e3q denote nonzero leading entries). Then the equation

c1r1+c2r2++ckrk=0

implies that

c1e11=0,c1e1p+c2e2p=0,c1e1q+c2e2q+c3e3q=0,

and so forth. We can therefore conclude in turn that c1=0, c2=0, , ck=0. Thus the row vectors r1,r2,,rk are linearly independent, and hence we have the following result.

We investigate the row space of an arbitrary matrix A by reducing A to an echelon matrix E. The following theorem then guarantees that A and E have the same row space. Recall that two matrices are (row) equivalent provided that each can be transformed to the other by elementary row operations.

Theorems 1 and 2 together provide a routine procedure for finding a basis for the row space of a given matrix, thereby determining its row rank.

Example 2

To find a basis for the row space of the matrix

(2)A=[12132349072351822835],

we reduce it to its echelon form

(3)E=[12132013540001700000].

Then the nonzero row vectors v1=(1, 2, 1, 3, 2), v2=(0, 1,3, 5,4), and v3=(0,0,0,1,7) form a basis for the row space of A. Thus Row(A) is a 3-dimensional subspace of R5 and the row rank of A is 3.

In Example 2, note that Algorithm 1 does not tell us that the first three row vectors of the matrix A itself are linearly independent. We know that some three row vectors of A span Row(A), but without further investigation we do not know which three.

Column Space and Column Rank

Now we turn our attention from row vectors to column vectors. Given an m×n matrix A=[aij], the column vectors of A are the n vectors

(4)c1=[a11a21am1],c2=[a12a22am2], ,cn=[a1na2namn]

in Rm. The subspace of Rm spanned by the n column vectors c1, c2, , cn is called the column space Col(A) of the matrix A. The dimension of the space Col(A) is called the column rank of the matrix A.

Example 3

Consider the 4×5 echelon matrix

(5)E=[12132013540001700000].

Its five column vectors c1, c2, c3, c4, c5 all lie in the subspace R3={(x1, x2, x3, 0)} of R4. The column vectors that contain the leading entries in the nonzero rows of E are

(6)c1=[1000],c2=[2100],andc4=[3510].

Note that

a1c1+a2c2+a4c4=(a1+2a2+3a4, a2+5a4, a4, 0).

Hence a1c1+a2c2+a4c4=0 readily implies that a1=a2=a4=0. Thus the vectors c1, c2, and c4 are linearly independent and therefore form a basis for R3={(x1, x2, x3, 0)}. Hence Col(E) is a 3-dimensional subspace of R4 and the column rank of E is 3.

To find the column rank of an arbitrary m×n matrix A, we begin in the same way as when finding the row rank. First we use elementary row operations to reduce A to an echelon matrix E. But the relationship between the column spaces of A and E is far more subtle than between their row spaces; the reason is that elementary row operations generally do not preserve column spaces. (See Problem 36.)

To analyze this situation, let us denote by c1, c2, , cn the column vectors of the original matrix A, and by c1, c2, , cn the column vectors of the echelon matrix to which we have reduced A. Suppose that E has k nonzero rows. Then each column vector of E has the form cj=(, , , 0, , 0) with mk final zeros. Hence the column space Col(E) is contained in Rk (considered as the subspace of Rm for which xk+1==xm=0).

We are particularly interested in those columns of E that contain the nonzero leading entries in the nonzero rows of E. These k columns are called the pivot columns of E. The echelon matrix E has the general form

(7)E=[d10d2000d30000dk00000000000000],

where d1, d2, , dk are the nonzero leading entries. Hence the k pivot column vectors of E look like

(8)[d100000],[d20000],[d3000], , [dk00].

Because of the upper triangular pattern visible here, it should be apparent that the argument in Example 3 can be used to show that these k pivot vectors form a basis for Col(E). Thus the column rank of the echelon matrix E is equal to the number k of its nonzero rows, and hence is equal to its row rank.

Now comes the subtle part. We want to show that the column rank of the original matrix A also is k. Note first that the homogeneous linear systems

(9)Ax=0andEx=0

have the same solution set because the matrices A and E are equivalent. If x=(x1, x2, , xn), then the left-hand sides in (9) are linear combinations of column vectors:

(10)Ax=x1c1+x2c2++xncn,Ex=x1c1+x1c2++xncn.

Because x satisfies either both or neither of the systems in (9), it follows that

x1c1+x2c2++xncn=0

if and only if

x1c1+x1c2++xncn=0.

Hence every linear dependence between column vectors of E is mirrored in a linear dependence with the same coefficients between the corresponding column vectors of A. In particular, because the k pivot column vectors of E are linearly independent, it follows that the corresponding k column vectors of A are linearly independent. And because the k pivot column vectors of E span Col(E), it follows that the corresponding k column vectors of A span Col(A). Thus the latter k vectors form a basis for the column space of A, and the column rank of A is k. Consequently we have established the following method for finding a basis for the column space of a given matrix.

Remark

Since the pivot column vectors of the matrix A—that is, the column vectors corresponding to the pivot column vectors of the echelon matrix E—form a basis for the column space of A, it follows that every non-pivot column vector of A is a linear combination of its pivot column vectors.

Example 4

Consider again the 4×5 matrix

A=[12132349072351822835]

of Example 2, in which we reduced it to the echelon matrix

E=[12132013540001700000].

The pivot columns of E are its first, second, and fourth columns, so the 3-dimensional column space of A has a basis consisting of its first, second, and fourth column vectors c1=(1, 3, 2, 2), c2=(2, 4, 3, 2), and c4=(3, 0, 1, 3).

The fact that Algorithm 2 provides a basis for Col(A) consisting of column vectors of A itself (rather than the echelon matrix E) enables us to apply it to the problem of extracting a maximal linearly independent subset from a given set of vectors in Rn.

Example 5

Find a subset of the vectors v1=(1, 1, 2, 2), v2=(3, 4, 1, 2), v3=(0, 1, 7, 4), and v4=(5, 7, 4,2) that forms a basis for the subspace W of R4 spanned by these four vectors.

Solution

It is not clear at the outset whether W is 2-dimensional, 3-dimensional, or 4-dimensional. To apply Algorithm 2, we arrange the given vectors as the column vectors of the matrix

A=[1305141721742242],

which reduces readily to the echelon matrix

E=[1305011200000000].

The pivot columns of E are its first two columns, so the first two column vectors v1=(1, 1, 2, 2) and v2=(3, 4, 1, 2) of A form a basis for the column space W. In particular, we see that W is 2-dimensional. (You can confirm that v3=3v1+v2 and v4=v1+2v2.)

Rank and Dimension

We have seen that the row rank and the column rank of an echelon matrix E both equal the number of rows of E that are not all zeros. But if A is any matrix that reduces by row operations to the echelon matrix E, then—according to Algorithm 1—the row rank of A is equal to the row rank of E, while Algorithm 2 implies that the column rank of A is equal to the column rank of E. We therefore have the following fundamental result.

For instance, if A is a 5×7 matrix and Row(A) is a 3-dimensional subspace of R7, then Theorem 3 tells us that Col(A) must be a 3-dimensional subspace of R5.

To see what an extraordinary theorem this is, think of a random 11×17 matrix A, possibly printed by a computer that uses a random number generator to produce independently the 187 entries in A. If it turns out that precisely 9 (no more) of the 11 row vectors of A are linearly independent, then Theorem 3 implies that precisely 9 (no more) of the 17 column vectors of A are linearly independent.

The common value of the row rank and the column rank of the matrix A is simply called the rank of A and is denoted by rank(A). The rank of A provides a meaning for the number of irredundant equations in the homogeneous linear system

(11)Ax=0.

Indeed, the individual scalar equations in (11) correspond to the column vectors of the transpose matrix AT, whose rank r equals that of A (Problem 25). Then the r columns vectors of AT that form a basis for Col(AT) correspond to the r equations in (11) that are irredundant; the remaining equations are linear combinations of these irredundant ones and hence are redundant.

The solution space of the homogeneous system Ax=0 is sometimes called the null space of A, denoted by Null(A). Note that if A is an m×n matrix, then Null(A) and Row(A) are subspaces of Rn, whereas Col(A) is a subspace of Rm. If r=rank(A), then the r column vectors of A that form a basis for Col(A) correspond to the r leading variables in a solution of Ax=0 by Gaussian elimination. Moreover, we know from the algorithm in Section 4.4 that the dimension of the solution space Null(A) is equal to the number nr of free variables. We therefore obtain the important identity

(12)rank(A)+dim Null(A)=n

for any m×n matrix A. For instance, if A is the matrix of Example 4 with rank 3 and n=5, then the dimension of the null space of A is 53=2.

For another typical application of Equation (12), consider a homogeneous system of five linear equations in seven unknowns. If the 5×7 coefficient matrix of the system has rank 3, so only 3 of the 5 equations are irredundant, then (because n=7) the system has 73=4 linearly independent solutions.

If A is an m×n matrix with rank m, then Equation (12) implies that the system Ax=0 has nm linearly independent solutions. Thus rank(A)=m is the condition under which the “conventional wisdom” about the relation between the numbers of equations, unknowns, and solutions is valid.

Nonhomogeneous Linear Systems

It follows immediately from Equation (10) and the definition of Col(A) that the nonhomogeneous linear system Ax=b is consistent if and only if the vector b is in the column space of A. In the problems below we outline some of the applications to nonhomogeneous systems of the results in this section.

If we can find a single particular solution x0 of the nonhomogeneous system

(13)Ax=b,

then the determination of its solution set reduces to solving the associated homogeneous system

(14)Ax=0.

For then Problem 29 in Section 4.2 implies that the solution set of (13) is the set of all vectors x of the form

x=x0+xh,

where xh is a solution of (14). Hence if x1, x2, , xr is a basis for the solution space of (14), then the solution set of the nonhomogeneous system is the set of all vectors of the form

x=x0+c1x1+c2x2++crxr.

We may describe this solution set as the translate by the vector x0 of the r-dimensional solution space of the corresponding homogeneous system.

4.5 Problems

In Problems 1–12, find both a basis for the row space and a basis for the column space of the given matrix A.

  1. [123159252]

  2. [524211415]

  3. [1437211712311]

  4. [13952141113313]

  5. [11113134251112]

  6. [1492226327163]

  7. [1117145161331325423]

  8. [1235149213712263]

  9. [13392748275122832]

  10. [12313134362243521323]

  11. [11331237822378331754]

  12. [11330102112378124067]

In Problems 13–16, a set S of vectors in R4 is given. Find a subset of S that forms a basis for the subspace of R4 spanned by S.

  1. v1=(1, 3, 2, 4), v2=(2, 1, 3, 2), v3=(5, 1, 4, 8)

  2. v1=(1, 1, 2, 3), v2=(2, 3, 4, 1), v3=(1, 1, 2, 1), v4=(4, 1, 8, 7)

  3. v1=(3, 2, 2, 2), v2=(2, 1, 2, 1), v3=(4, 3, 2, 3), v4=(1, 2, 3, 4)

  4. v1=(5, 4, 2, 2), v2=(3, 1, 2, 3), v3=(7, 7, 2, 1), v4=(1, 1, 2, 4), v5=(5, 4, 6, 7)

Let S={v1, v2, , vk} be a basis for the subspace W of Rn. Then a basis T for Rn that contains S can be found by applying the method of Example 5 to the vectors

v1, v2, , vk, e1, e2, , en.

Do this in Problems 17–20.

  1. Find a basis T for R3 that contains the vectors v1=(1, 2, 2) and v2=(2, 3, 3).

  2. Find a basis T for R3 that contains the vectors v1=(3, 2, 1) and v2=(2, 2, 1).

  3. Find a basis T for R4 that contains the vectors v1=(1, 1, 1, 1) and v2=(2, 3, 3, 3).

  4. Find a basis T for R4 that contains the vectors v1=(3, 2, 3, 3) and v2=(5, 4, 5, 5).

Given a homogeneous system Ax=0 of (scalar) linear equations, we say that a subset of these equations is irredundant provided that the corresponding column vectors of the transpose AT are linearly independent. In Problems 21–24, extract from each given system a maximal subset of irredundant equations.

  1. x13x2+2x3=02x1+3x2+2x3=04x13x2+6x3=0

  2. 3x1+2x2+2x3+2x4=02x1+3x2+3x3+3x4=08x1+7x2+7x3+7x4=0

  3. x1+2x2x3+2x4=03x1x2+3x3+x4=05x1+3x2+x3+5x4=02x1+5x3+4x4=0

  4. 3x1+2x2+2x3=02x1+3x2+3x3=07x1+8x2+8x3=08x1+7x2+7x3=05x1+6x2+5x3=0

  5. Explain why the rank of a matrix A is equal to the rank of its transpose AT.

  6. Explain why the n×n matrix A is invertible if and only if its rank is n.

Problems 27 through 32 explore the properties of nonhomogeneous systems.

  1. Let A be a 3×5 matrix whose three row vectors are linearly independent. Prove that, for each b in R3, the nonhomogeneous system Ax=b has a solution.

  2. Let A be a 5×3 matrix that has three linearly independent row vectors. Suppose that b is a vector in R5 such that the nonhomogeneous system Ax=b has a solution. Prove that this solution is unique.

  3. Let A be an m×n matrix with m<n. Given b in Rm, note that any solution of Ax=b expresses b as a linear combination of the column vectors of A. Then prove that no such solution is unique.

  4. Given the m×n matrix A with m>n, show that there exists a vector b in Rm such that the system Ax=b has no solution.

  5. Existence of Solutions Let A be an m×n matrix. Prove that the system Ax=b is consistent for every b in Rm if and only if the rank of A is equal to m.

  6. Uniqueness of Solutions Let A be an m×n matrix and suppose that the system Ax=b is consistent. Prove that its solution is unique if and only if the rank of A is equal to n.

  7. Prove that the pivot column vectors in (8) are linearly independent.

  8. Let A be a matrix with rank r, and suppose that A can be reduced to echelon form without row interchanges. Show that the first r row vectors of A are linearly independent.

  9. Deduce from Theorem 3 in Section 4.3 that the rank of the matrix A is the largest integer r such that A has a nonsingular r×r submatrix.

  10. Let A be the 3×2 matrix whose columns are the unit basis vectors i=(1, 0, 0) and j=(0, 1, 0). If B is the row-equivalent matrix obtained by adding the first and second rows of A to its third row, show that A and B do not have the same column space.

4.6 Orthogonal Vectors in Rn

In this section we show that the geometrical concepts of distance and angle in n-dimensional space can be based on the definition of the dot product of two vectors in Rn. Recall from elementary calculus that the dot product uv of two 3-dimensional vectors u=(u1, u2, u3) and v=(v1, v2, v3) is (by definition) the sum

uv=u1v1+u2v2+u3v3

of the products of corresponding scalar components of the two vectors.

Similarly, the dot product uv of two n-dimensional vectors u=(u1, u2, , un) and v=(v1, v2, , vn) is defined by

(1)uv=u1v1+u2v2++unvn

(with one additional scalar product term for each additional dimension). And just as in R3, it follows readily from the formula in (1) that if u, v, and w are vectors in Rn and c is a scalar, then

(2)
(3)
(4)
(5)
uv=vu(symmetry)u(v+w)=uv+uw(distributivity)(cu)v=c(uv)(homogeneity)uu0;uu=0if and only ifu=0.}(positivity)

Therefore, the dot product in Rn is an example of an inner product.

The dot product on Rn is sometimes called the Euclidean inner product, and with this inner product Rn is sometimes called Euclidean n-dimensional space. We can use any of the notations in

uv=u, v=uTv

for the dot product of the two n×1 column vectors u and v. (Note in the last expression that uT=(u1, u2, , un)T is the 1×n row vector with the indicated entries, so the 1×1 matrix product uTv is simply a scalar.) Here we will ordinarily use the notation uv.

The length |u| of the vector u=(u1, u2, , un) is defined as follows:

(6)|u|=(uu)=(u12+u22++un2)1/2.

Note that the case n=2 is a consequence of the familiar Pythagorean formula in the plane.

Theorem 1 gives one of the most important inequalities in mathematics. Many proofs are known, but none of them seems direct and well motivated.

The Cauchy-Schwarz inequality enables us to define the angle θ between the nonzero vectors u and v. (See Figure 4.6.1.) Division by the positive number |u||v| in (7) yields |uv|/(|u||v|)1, so

(8)1uv|u||v|+1.

FIGURE 4.6.1.

The angle θ between the vectors u and v.

Hence there is a unique angle θ between 0 and π radians, inclusive (that is, between 0° and 180°), such that

(9)cos θ=uv|u||v|.

Thus we obtain the same geometric interpretation

(10)uv=|u||v|cos θ

of the dot product in Rn as one sees (for 3-dimensional vectors) in elementary calculus textbooks—for instance, see Section 11.2 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008.

On the basis of (10) we call the vectors u and v orthogonal provided that

(11)uv=0.

If u and v are nonzero vectors this means that cos θ=0, so θ=π/2(90°). Note that u=0 satisfies (11) for all v, so the zero vector is orthogonal to every vector.

Example 1

Find the angle θn in Rn between the x1-axis and the line through the origin and the point (1, 1, …, 1).

Solution

We take u=(1, 0, 0, , 0) on the x1-axis and v=(1, 1, , 1). Then |u|=1, |v|=n, and uv=1, so the formula in (9) gives

cos θn=uv|u||v|=1n.

For instance, if

n=3,thenθ3=cos1(13)0.9553(55°);n=4,thenθ4=cos1(14)1.0472(60°);n=5,thenθ5=cos1(15)1.1071(63°);n=100,thenθ100=cos1(110)1.4706(84°).

It is interesting to note that θn increases as n increases. Indeed, θn approaches cos1(0)=π/2(90°) as n increases without bound (so that 1/n approaches zero).

In addition to angles, the dot product provides a definition of distance in Rn. The distance d(u, v) between the points (vectors) u=(u1, u2, , un) and v=(v1, v2, , vn) is defined to be

(12)d(u,v)=|uv|=[(u1v1)2+(u2v2)2++(unvn)2]1/2.

Example 2

The distance between the points u=(1, 1, 2, 3, 5) and v=(4, 3, 4, 5, 9) in R5 is

|uv|=32+42+62+22+42=81=9.

The triangle inequality of Theorem 2 relates the three sides of the triangle shown in Figure 4.6.2.

FIGURE 4.6.2.

The “triangle” of the triangle inequality.

The vectors u and v are orthogonal if and only if uv=0, so line (14) in the proof of the triangle inequality yields the fact that the Pythagorean formula

(14)|u+v|2=|u|2+|v|2

holds if and only if the triangle with “adjacent side vectors” u and v is a right triangle with hypotenuse vector u+v (see Figure 4.6.3).

FIGURE 4.6.3.

A right triangle in Rn.

The following theorem states a simple relationship between orthogonality and linear independence.

In particular, any set of n mutually orthogonal nonzero vectors in Rn constitutes a basis for Rn. Such a basis is called an orthogonal basis. For instance, the standard unit vectors e1,e2, , en form an orthogonal basis for Rn.

Orthogonal Complements

Now we want to relate orthogonality to the solution of systems of linear equations. Consider the homogeneous linear system

(15)Ax=0

of m equations in n unknowns. If v1, v2, , vm are the row vectors of the m×n coefficient matrix A, then the system looks like

Consequently, it is clear that x is a solution vector of Ax=0 if and only if x is orthogonal to each row vector of A. But in the latter event x is orthogonal to every linear combination of row vectors of A because

x(c1v1+c2v2++cmvm)=c1xv1+c2xv2++cmxvm=(c1)(0)+(c2)(0)++(cm)(0)=0.

Thus we have shown that the vector x in Rn is a solution vector of Ax=0 if and only if x is orthogonal to each vector in the row space Row(A) of the matrix A. This situation motivates the following definition.

If u1 and u2 are vectors in V, v is in V, and c1 and c2 are scalars, then

(c1u1+c2u2)v=c1u1v+c2u2v=(c1)(0)+(c2)(0)=0.

Thus any linear combination of vectors in V is orthogonal to every vector in V and hence is a vector in V. Therefore the orthogonal complement V of a subspace V is itself a subspace of Rn. The standard picture of two complementary subspaces V and V consists of an orthogonal line and plane through the origin in R3 (see Fig. 4.6.4). The proofs of the remaining parts of Theorem 4 are left to the problems.

FIGURE 4.6.4.

Orthogonal complements.

In our discussion of the homogeneous linear system Ax=0 in (16), we showed that a vector space lies in the null space Null(A) of A—that is, in the solution space of Ax=0—if and only if it is orthogonal to each vector in the row space of A. In the language of orthogonal complements, this proves Theorem 5.

Now suppose that a subspace V of Rn is given, with v1, v2, , vm a set of vectors that span V. For instance, these vectors may form a given basis for V. Then the implication in (17) provides the following algorithm for finding a basis for the orthogonal complement V of V.

  1. Let A be the m×n matrix with row vectors v1, v2, , vm.

  2. Reduce A to echelon form and use the algorithm of Section 4.4 to find a basis {u1, u2, , uk} for the solution space Null(A) of Ax=0. Because V=Null(A), this will be a basis for the orthogonal complement of V.

Example 3

Let V be the 1-dimensional subspace of R3 spanned by the vector v1=(1, 3, 5). Then

A=[135]

and our linear system Ax=0 consists of the single equation

x13x2+5x3=0.

If x2=s and x3=t, then x1=3s5t. With s=1 and t=0 we get the solution vector u1=(3, 1, 0), whereas with s=0 and t=1 we get the solution vector u2=(5, 0, 1). Thus the orthogonal complement V is the 2-dimensional subspace of R3 having u1=(3, 1, 0) and u2=(5, 0, 1) as basis vectors.

Example 4

Let V be the 2-dimensional subspace of R5 that has v1=(1, 2, 1, 3, 3) and v2=(2, 5, 6, 10, 12) as basis vectors. The matrix

A=[121332561012]

with row vectors v1 and v2 has reduced echelon form

E=[1075901446].

Hence the solution space of Ax=0 is described parametrically by

x3=r,x4=s,x5=t,x2=4r+4s+6tx1=7r5s9t.

Then the choice

r=1,s=0,t=0yieldsu1=(7, 4, 1, 0, 0);r=0,s=1,t=0yieldsu2=(5, 4, 0, 1, 0);r=0,s=0,t=1yieldsu3=(9, 6, 0, 0, 1).

Thus the orthogonal complement V is the 3-dimensional subspace of R5 with basis {u1, u2, u3}.

Observe that dimV+dimV=3 in Example 3, but dimV+dimV=5 in Example 4. It is no coincidence that in each case the dimensions of V and V add up to the dimension n of the Euclidean space containing them. To see why, suppose that V is a subspace of Rn and let A be an m×n matrix whose row vectors span V. Then Equation (12) in Section 4.5 implies that

rank(A)+dimNull(A)=n.

But

dimV=dimRow(A)=rank(A)

and

dimV=dimNull(A)

by Theorem 5, so it follows that

(17)dimV+dimV=n.

Moreover, it should be apparent intuitively that if

{v1, v2, , vm} is a basis for V

and

{u1, u2, , uk} is a basis for V,

then

{v1, v2, , vm, u1, u2, , uk} is a basis for Rn.

That is, the union of a basis for V and a basis for V is a basis for Rn. In Problem 34 of this section we ask you to prove that this is so.

4.6 Problems

In Problems 1–4, determine whether the given vectors are mutually orthogonal.

  1. v1=(2, 1, 2, 1), v2=(3, 6, 1, 2), v3=(3, 1, 5, 5)

  2. v1=(3, 2, 3, 4), v2=(6, 3, 4, 6), v3=(17, 12, 21, 3)

  3. v1=(5, 2, 4, 1), v2=(3, 5, 1, 1), v3=(3, 0, 8, 17)

  4. v1=(1, 2, 3, 2, 1), v2=(3, 2, 3, 6, 4), v3=(6, 2, 4, 1, 4)

In Problems 5–8, the three vertices A, B, and C of a triangle are given. Prove that each triangle is a right triangle by showing that its sides a, b, and c satisfy the Pythagorean relation a2+b2=c2.

  1. A(6, 6, 5, 8), B(6, 8, 6, 5), C(5, 7, 4, 6)

  2. A(3, 5, 1, 3), B(4, 2, 6, 4), C(1, 3, 4, 2)

  3. A(4, 5, 3, 5, 1), B(3, 4, 1, 4, 4), C(1, 3, 1, 3, 1)

  4. A(2, 8, 3, 1, 2), B(2, 5, 6, 2, 12), C(5, 3, 2, 3, 5)

  1. 9–12. Find the acute angles (in degrees) of each of the right triangles of Problems 5–8, respectively.

In Problems 13–22, the given vectors span a subspace V of the indicated Euclidean space. Find a basis for the orthogonal complement V of V.

  1. v1=(1, 2, 3)

  2. v1=(1, 5, 3)

  3. v1=(1, 2, 3, 5)

  4. v1=(1, 7, 6, 9)

  5. v1=(1, 3, 2, 4), v2=(2, 7, 7, 3)

  6. v1=(1, 3, 3, 5), v2=(2, 5, 9, 3)

  7. v1=(1, 2, 5, 2, 3), v2=(3, 7, 11, 9, 5)

  8. v1=(2, 5, 5, 4, 3), v2=(3, 7, 8, 8, 8)

  9. v1=(1, 2, 3, 1, 3), v2=(1, 3, 4, 3, 6), v3=(2, 2, 4, 3, 5)

  10. v1=(1, 1, 1, 1, 3), v2=(2, 3, 1, 4, 7), v3=(5, 3, 7, 1, 5)

  11. Prove: For arbitrary vectors u and v,

    1. |u+v|2+|uv|2=2|u|2+2|v|2;

    2. |u+v|2|uv|2=4uv.

  12. Given mutually orthogonal vectors v1, v2, , vk, show that

    |v1+v2++vk|2=|v1|2+|v2|2++|vk|2.
  13. Suppose that A, B, and C are the unit points on distinct coordinate axes in Rn (for example, perhaps A=e1, B=e2, and C=e3). Show that ABC is an equilateral triangle.

  14. Prove that |uv|=|u||v| if and only if the vectors u and v are linearly dependent.

In Problems 27–29, V denotes a subspace of Rn.

  1. Show that the only vector that lies in both V and V is the zero vector.

  2. Prove that if W=V then W=V.

  3. Let S be a spanning set for V. Show that the vector u is in V if and only if u is orthogonal to every vector in S.

  4. Suppose that the vectors u and v are orthogonal and that u+v=0. Show that u=v=0.

  5. Let S and T be two sets of vectors in Rn such that each vector in S is orthogonal to every vector in T. Prove that each vector in span(S) is orthogonal to every vector in span(T).

  6. Let S={u1, u2} and T={v1, v2} be linearly independent sets of vectors such that each ui in S is orthogonal to every vector vj in T. Then use the results of Problems 30 and 31 to show that the four vectors u1, u2, v1, v2 are linearly independent.

  7. Let S={u1, u2, , uk} and T={v1, v2, , vm} be linearly independent sets of vectors such that each ui in S is orthogonal to every vector vj in T. Then generalize Problem 32 to show that the k+m vectors u1, u2, , uk, v1, v2, , vm are linearly independent.

  8. Deduce from the result of Problem 33 that if V is a subspace of Rn, then the union of a basis for V and a basis for V is a basis for Rn.

  9. Let A be an m×n matrix and b a vector in Rm. Combine the observations listed below to prove that the nonhomogeneous linear system Ax=b is consistent if and only if b is orthogonal to the solution space of the homogeneous system ATy=0.

    1. Ax=b is consistent if and only if b is in Col(A).

    2. b is in Col(A) if and only if b is orthogonal to {Col(A)}={Row(AT)}.

    3. {Row(AT)}=Null(AT).

4.7 General Vector Spaces

In the previous six sections of this chapter, almost all the specific vector spaces appearing in our examples and problems have been vector spaces of n-tuples of real numbers. We have thus confined our attention largely to Euclidean spaces and their subspaces. In this section we discuss examples of some other types of vector spaces that play important roles in various branches of mathematics and their applications.

Example 1

Given fixed positive integers m and n, let Mmn denote the set of all m×n matrices with real number entries. Then Mmn is a vector space with the usual operations of addition of matrices and multiplication of matrices by scalars (real numbers). That is, these operations satisfy properties (a)–(h) in the definition of a vector space (Section 4.2). In particular, the zero element in Mmn is the m×n matrix 0 matrix whose elements are all zeros, and the negative A of the matrix A is (as usual) the matrix whose elements are the negatives of the corresponding entries of A.

Given positive integers i and j with 1im and 1jn, let Eij denote the m×n matrix whose only nonzero entry is the number 1 in the ith row and the jth column. Then it should be clear that the mn elements {Eij} of Mmn form a basis for Mmn, and so Mmn is a finite-dimensional vector space of dimension mn.

For instance, in the case of the vector space M22 of all 2×2 matrices, these alleged basis elements are the four matrices

E11=[1000],E12=[0100],E21=[0010],andE22=[0001].

Then any matrix A in M22 can be expressed as

A=[abcd]=aE11+bE12+cE21+dE22,

so the set {E11, E12, E21, E22} spans M22. Moreover,

aE11+bE12+cE21+dE22=[abcd]=[0000]

implies immediately that a=b=c=d=0, so the set {E11, E12, E21, E22} is also linearly independent and therefore forms a basis for M22. Thus, we have shown that M22 is a 4-dimensional vector space.

Example 2

Let C denote the subset of M22 consisting of all 2×2 matrices of the form

(1)[abba].

Obviously the sum of any two such matrices is such a matrix, as is any scalar multiple of such a matrix. Thus C is closed under the operations of addition of matrices and multiplication by scalars and is therefore a subspace of M22.

To determine the dimension of the subspace C, we consider the two special matrices

(2)Re=[1001]andIm=[0110]

in C. Here Re is another notation for the 2×2 identity matrix I, whereas the matrix Im is not so familiar. The matrices Re and Im are linearly independent—obviously neither is a scalar multiple of the other—and

(3)[abba]=aRe+bIm.

so these two matrices span C. Thus {Re, Im} is a basis for C, and so C is a 2-dimensional subspace of the 4-dimensional vector space M22 of 2×2 matrices.

Now observe that

(4)(Im)2=[0110][0110]=[1001]=Re.

It follows that the matrix product of any two elements aRe+bIm and cRe+dIm of C is given by

(5)(aRe+bIm)(cRe+dIm)=ac(Re)2+bcImRe+adReIm+bd(Im)2=(acbd)Re+(ad+bc)Im.

This explicit formula for the product of two matrices in C shows that, in addition to being closed under matrix addition and multiplication by scalars, the subspace C of M22 also is closed under matrix multiplication.

If you feel that the idea of a space of matrices being closed under matrix multiplication is a rather complex one, you are correct! Indeed, the vector space C of Example 2 can be taken as a “model” for the set of all complex numbers of the form a+bi, where a and b are real numbers and i=1 denotes the “imaginary” square root of 1 (so that i2=1). Because (Im)2=Re, the matrix Im plays the role of i and the matrix

[abba]=aRe+bIm

corresponds to the complex number a+bi. And the matrix multiplication formula in (5) is then analogous to the formula

(a+bi)(c+di)=(acbd)+(ad+bc)i

for the multiplication of complex numbers. In addition to providing a concrete interpretation of complex numbers in terms of real numbers, the matrix model C provides a convenient way to manipulate complex numbers in computer programming languages in which matrix operations are “built in” but complex operations are not.

Function Spaces

In Example 1 of Section 4.2 we introduced the vector space F of all real-valued functions defined on the real line R. If f and g are elements of F and c is a scalar, then

(6)(f+g)(x)=f(x)+g(x)

and

(cf)(x)=c(f(x))

for all x in R. The zero element in F is the function 0 such that 0(x)=0 for all x. We frequently refer to a function in F by simply specifying its formula. For instance, by “the function x2” we mean that function whose value at each x in R is x2.

The functions f1,f2, ,fk in F are linearly dependent provided that there exist scalars c1,c2, ,ck, not all zero, such that

c1f1(x)+c2f2(x)++ckfk(x)=0

for all x in R. We often can determine whether two given functions are linearly dependent simply by observing whether one is a scalar multiple of the other.

Example 3

The functions ex and e2x are linearly independent because either of the equations ex=ae2x or e2x=bex would imply that e3x is a constant, which obviously is not so. By contrast, the functions sin 2x and sin x cos x are linearly dependent, because of the trigonometric identity sin 2x=2sin xcos x. The three functions 1, cos2 x, and sin2 x are linearly dependent, because the fundamental identity cos2 x+sin2 x=1 can be written in the form

(1)(1)+(1)(cos2 x)+(1)(sin2 x)=0.

A subspace of F is called a function space. An example of a function space is the vector space P of all polynomials in F (Example 4 in Section 4.4). Recall that a function p(x) in F is called a polynomial of degree n0 if it can be expressed in the form

(7)p(x)=a0+a1x+a2x2++anxn

with an0. Polynomials are added and multiplied by scalars in the usual manner—by collecting coefficients of like powers of x. Clearly any linear combination of polynomials is a polynomial, so P is, indeed, a subspace of F. (Recall the subspace criterion of Theorem 1 in Section 4.2.)

Example 4

Given n0, denote by Pn the set of all polynomials of degree at most n. That is, Pn consists of all polynomials of degrees 0, 1, 2, …, n. Any linear combination of two polynomials of degree at most n is again a polynomial of degree at most n, so Pn is a subspace of P (and of F). The formula in (7) shows that the n+1 polynomials

(8)1, x, x2, x3, , xn

span Pn. To show that these “monomials” are linearly independent, suppose that

(9)c0+c1x+c2x2++cnxn=0

for all x in R. If a0, a1, , an are fixed distinct real numbers, then each ai satisfies Eq. (9).

In matrix notation this means that

(10)[1a0a02a0n1a1a12a1n1a2a22a2n1anan2ann][c0c1c2cn]=[0000].

The coefficient matrix in (10) is a Vandermonde matrix V, and Problems 61–63 in Section 3.6 imply that any Vandermonde matrix is nonsingular. Therefore the system Vc=0 has only the trivial solution c=0, so it follows from (10) that c0=c1=c2==cn=0. Thus we have proved that the n+1 monomials 1, x, x2, , xn are linearly independent and hence constitute a basis for Pn. Consequently Pn is an (n+1)-dimensional vector space.

The fact that the monomials in (8) are linearly independent implies that the coefficients in a polynomial are unique. That is, if

a0+a1x++anxn=b0+b1x++bnxn

for all x, then a0=b0, a1=b1, , and an=bn. (See the remark that follows the definition of linear independence in Section 4.3.) This identity principle for polynomials is often used without proof in elementary algebra. A typical application is the method of partial-fraction decomposition illustrated in Example 5.

Example 5

Find constants A, B, and C such that

(11)6x(x1)(x+1)(x+2)=Ax1+Bx+1+Cx+2

for all x (other than x=1, 1, or 2).

Solution

Multiplication of each side of the equation in (11) by the denominator on the left-hand side yields

6x=A(x+1)(x+2)+B(x1)(x+2)+C(x1)(x+1);6x=(2A2BC)+(3A+B)x+(A+B+C)x2.

Then the identity principle for polynomials yields the linear equations

2A2BC=03A+B=6A+B+C=0,

which we readily solve for A=1, B=3, and C=4. Therefore

6x(x1)(x+1)(x+2)=1x1+3x+14x+2

if x1, 1, 2.

Example 6

Show that the four polynomials

(12)1,x,3x21,and5x33x

form a basis for P3.

Solution

In order to show simultaneously that these four polynomials span P3 and are linearly independent, it suffices to see that every polynomial

(13)p(x)=b0+b1x+b2x2+b3x3

in P3 can be expressed uniquely as a linear combination

(14)c0(1)+c1(x)+c2(3x21)+c3(5x33x)=(c0c2)+(c13c3)x+(3c2)x2+(5c3)x3

of the polynomials in (12). But upon comparing the coefficients in (13) and (14), we see that we need only observe that the linear system

[1010010300300005][c0c1c2c3]=[b0b1b2b3]

obviously has a unique solution for c0, c1, c2, c3 in terms of b0, b1, b2, b3.

Because the vector space P of all polynomials contains n+1 linearly independent functions for every integer n0, it follows that P is an infinite-dimensional vector space. So too is any function space, such as F itself, that contains P. Another important function space is the set C(0) of all continuous functions on R; C(0) is a subspace of F because every linear combination of continuous functions is again a continuous function. Every polynomial is a continuous function, so C(0) contains P and is therefore an infinite-dimensional vector space. Similarly, the set C(k) of all functions in F that have continuous kth-order derivatives is an infinite-dimensional function space.

Solution Spaces of Differential Equations

In Section 1.5 we saw that the (now familiar) first-order differential equation

(15)y=ky(k constant)

has the general solution y(x)=Cekx. Thus the “solution space” of (15) is the set of all constant multiples of the single function ekx, and it is therefore the 1-dimensional function space with basis {ekx}.

In Section 5.1 we will see that the set of all solutions y(x) of a linear second-order differential equation of the form

(16)ay+by+cy=0

(with constant coefficients a, b, and c) is a 2-dimensional function space S, called the solution space of the differential equation. Here we will illustrate this general fact with three simple examples of second-order equations that can be solved in an elementary manner.

Example 7

With a=1 and b=c=0 in (16), we get the differential equation

(17)y=0.

If y(x) is any solution of this equation, so y(x)=0, then two integrations give first

y(x)=y(x) dx=(0) dx=A

and then

y(x)=y(x) dx=Adx=Ax+B,

where A and B are arbitrary constants of integration. Thus every solution of (17) is of the form y(x)=Ax+B, and clearly every such function is a solution of the differential equation. Thus the solution space of (17) is simply the 2-dimensional space P1 of linear polynomials generated by the basis {1,x}.

Example 8

With a=1, b=2, and c=0 in (16), we get the differential equation

(18)y2y=0.

If y(x) is any solution of this equation and if v(x)=y(x), then (18) says that

v(x)=2v(x).

But this is a familiar first-order differential equation with general solution v(x)=Ce2x. Then integration gives

y(x)=y(x) dx=v(x) dx=Ce2x dx=12Ce2x+A.

Thus every solution of (18) is of the form y(x)=A+Be2x (where B=12C), and you should show by substitution in (18) that every function of this form is a solution of the differential equation. Therefore, the solution space of (18) is the 2-dimensional function space that is generated by the basis {1,e2x}.

Example 9

With a=1, b=0, and c=1 in (16), we get the differential equation

(19)yy=0.

If y(x) is any solution of this equation and if v(x)=y(x), then (19) says that

y=dvdx=dvdydydx=vdvdy=y;

so v dv=y dy. Integration therefore gives

(20)12v2=12y2+C,sov2=y2±a2,

where the final sign depends on the sign of C. The final result is independent of this choice (Problem 27); we’ll take the negative sign for the sake of illustration. Since v=dy/dx, it then follows from (20) that

(21)x=±dyy2a2.

Here again the final result is independent of the choice of sign; this time we’ll take the positive sign. Then the substitution y=au and the standard integral

duu21=cosh1 u+C

(see Section 7.6 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008) yield

x=cosh1ya+b,

so finally,

y=a cosh(xb)=a(cosh xcosh bsinh xsinh b),

which can be written as

(22)y(x)=A cosh x+B sinh x,

with A=a cosh b and B=a sinh b. Thus every solution of the differential equation yy=0 in (19) is of the form y(x)=A cosh x+B sinh x, and you should show by substitution in (19) that every function of this form is a solution of the differential equation. Therefore, the solution space S of (19) is the 2-dimensional function space that is generated by the basis {cosh x,sinh x}. If we substitute into (22) the relations

cosh x=ex+ex2,sinh x=exex2

that define the hyperbolic cosine and sine functions in terms of exponentials, we see that every solution of yy=0 can also be written in the form of the linear combination

y(x)=Cex+Dex.

Consequently, {ex, ex} is a second basis for the 2-dimensional solution space S.

In Section 5.3 we will learn a much quicker way to solve linear second-order differential equations like the equation yy=0 of Example 9 and the equation y+y=0 of Problem 28.

4.7 Problems

In Problems 1–4, determine whether or not the indicated set of 3×3 matrices is a subspace of M33.

  1. The set of all diagonal 3×3 matrices.

  2. The set of all symmetric 3×3 matrices (that is, matrices A=[aij] such that aij=aji for 1i3, 1j3).

  3. The set of all nonsingular 3×3 matrices.

  4. The set of all singular 3×3 matrices.

In Problems 5–8, determine whether or not each indicated set of functions is a subspace of the space F of all real-valued functions on R.

  1. The set of all f such that f(0)=0.

  2. The set of all f such that f(x)0 for all x.

  3. The set of all f such that f(0)=0 and f(1)=1.

  4. The set of all f such that f(x)=f(x) for all x.

In Problems 9–12, a condition on the coefficients of a polynomial a0+a1x+a2x2+a3x3 is given. Determine whether or not the set of all such polynomials satisfying this condition is a subspace of the space P of all polynomials.

  1. a30

  2. a0=a1=0

  3. a0+a1+a2+a3=0

  4. a0, a1, a2, and a3 are all integers

In Problems 13–18, determine whether the given functions are linearly independent.

  1. sin x and cos x

  2. ex and xex

  3. 1+x, 1x, and 1x2

  4. 1+x, x+x2, and 1x2

  5. cos 2x, sin2 x, and cos2 x

  6. 2 cos x+3 sin x and 4 cos x+5 sin x

In Problems 19–22, use the method of Example 5 to find the constants A, B, and C in the indicated partial-fraction decompositions.

  1. x5(x2)(x3)=Ax2+Bx3

  2. 2x(x21)=Ax+Bx1+Cx+1

  3. 8x(x2+4)=Ax+Bx+Cx2+4

  4. 2x(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3

In Problems 23 and 24, use the method of Example 7 to find a basis for the solution space of the given differential equation. (It’s 3-dimensional in Problem 23 and 4-dimensional in Problem 24.)

  1. y=0

  2. y(4)=0

In Problems 25 and 26, use the method of Example 8 to find a basis for the 2-dimensional solution space of the given differential equation.

  1. y5y=0

  2. y+10y=0

  3. Take the positive sign in Eq. (20), and then use the standard integral

    duu2+1=sinh1 u+C

    to derive the same general solution y(x)=A cosh x+B sinh x given in Eq. (22).

  4. Use the method of Example 9 and the standard integral

    du1u2=sinh1 u+C

    to derive the general solution y(x)=A cos x+B sin x of the second-order differential equation y+y=0. Thus its solution space has basis {cos x,sin x}.

  5. Let V be the set of all infinite sequences {xn}={x1, x2, x3, } of real numbers. Let addition of elements of V and multiplication by scalars be defined as follows:

    {xn}+{yn}={xn+yn}

    and

    c{xn}={cxn}.
    1. Show that V is a vector space with these operations.

    2. Prove that V is infinite dimensional.

  6. Let V be the vector space of Problem 29 and let the subset W consist of those elements {xn} of V such that xn=xn1+xn2 for n2. Thus a typical element of W is the Fibonacci sequence

    {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, }.
    1. Show that W is a subspace of V.

    2. Prove that W is 2-dimensional.

Problems 31 through 33 develop a method of multiplying matrices whose entries are complex numbers.

  1. Motivated by Example 2, let us define a function or transformation T from the set of all complex numbers to the space of all 2×2 matrices of the form in (1) as follows. Given a complex number z=a+bi, let

    T(z)=[abba].
    1. Suppose that z1 and z2 are complex numbers and c1 and c2 are real numbers. Show that

      T(c1z1+c2z2)=c1T(z1)+c2T(z2).
    2. Show that

      T(z1z2)=T(z1)T(z2)

      for all complex numbers z1 and z2. Note the complex multiplication on the left in contrast with the matrix multiplication on the right.

    3. Prove that if z is an arbitrary nonzero complex number, then

      T(z1)={T(z)}1,

      where z1=1/z and {T(z)}1 is the inverse of the matrix T(z).

  2. Let A and B be 4×4 (real) matrices partitioned into 2×2 submatrices or “blocks”:

    A=[A11A12A21A22],B=[B11B12B21B22].

    Then verify that AB can be calculated in “blockwise” fashion:

    AB=[A11B11+A12B21A11B12+A12B22A21B11+A22B21A21B12+A22B22].
  3. Given a 2×2 matrix M=[zij] whose entries are complex numbers, let T(M) denote the 4×4 matrix of real numbers given in block form by

    T(M)=[T(z11)T(z12)T(z21)T(z22)].

    The transformation T on the right-hand side is the one we defined in Problem 31. Suppose that M and N are 2×2 complex matrices. Use Problems 31(b) and 32 to show that

    T(MN)=T(M)T(N).

    Hence one can find the product of the 2×2 complex matrices M and N by calculating the product of the 4×4 real matrices T(M) and T(N). For instance, if

    M=[1+i2+i1+2i1+3i]

    and

    N=[3i12i2i32i],

    then

    T(M)T(N)=[1121111212132131]×[3112132121321223]=[921122921110101471010714].

    Therefore,

    MN=[9+2i112i10+10i14+7i].

    If M is a nonsingular 2×2 complex matrix, it can be shown that T(M1)={T(M)}1. The n×n versions of these results are sometimes used to carry out complex matrix operations in computer languages that support only real arithmetic.

5 Higher-Order Linear Differential Equations

5.1 Introduction: Second-Order Linear Equations

In Chapters 1 and 2 we investigated first-order differential equations. We now turn to equations of higher order n2, beginning in this chapter with equations that are linear. The general theory of linear differential equations parallels the second-order case (n=2), which we outline in this initial section.

Recall that a second-order differential equation in the (unknown) function y(x) is one of the form

(1)G(x,y,y,y)=0.

This differential equation is said to be linear provided that G is linear in the dependent variable y and its derivatives y and y. Thus a linear second-order equation takes (or can be written in) the form

(2)A(x)y+B(x)y+C(x)y=F(x).

Unless otherwise noted, we will always assume that the (known) coefficient functions A(x), B(x), C(x), and F(x) are continuous on some open interval I (perhaps unbounded) on which we wish to solve this differential equation, but we do not require that they be linear functions of x. Thus the differential equation

exy+(cosx)y+(1+x)y=tan1x

is linear because the dependent variable y and its derivatives y and y appear linearly. By contrast, the equations

y=yyandy+3(y)2+4y3=0

are not linear because products and powers of y or its derivatives appear.

If the function F(x) on the right-hand side of Eq. (2) vanishes identically on I, then we call Eq. (2) a homogeneous linear equation; otherwise, it is nonhomogeneous. For example, the second-order equation

x2y+2xy+3y=cosx

is nonhomogeneous; its associated homogeneous equation is

x2y+2xy+3y=0.

In general, the homogeneous linear equation associated with Eq. (2) is

(3)A(x)y+B(x)y+C(x)y=0.

In case the differential equation in (2) models a physical system, the nonhomogeneous term F(x) frequently corresponds to some external influence on the system.

Remark

Note that the meaning of the term “homogeneous” for a second-order linear differential equation is quite different from its meaning for a first-order differential equation (as in Section 1.6). Of course, it is not unusual—either in mathematics or in the English language more generally—for the same word to have different meanings in different contexts.

A Typical Application

Linear differential equations frequently appear as mathematical models of mechanical systems and electrical circuits. For example, suppose that a mass m is attached both to a spring that exerts on it a force FS and to a dashpot (shock absorber) that exerts a force FR on the mass (Fig. 5.1.1). Assume that the restoring force FS of the spring is proportional to the displacement x of the mass from its equilibrium position and acts opposite to the direction of displacement. Then

FS=kx(with k>0)

FIGURE 5.1.1.

A mass–spring–dashpot system.

so FS<0 if x>0 (spring stretched) while FS>0 if x<0 (spring compressed). We assume that the dashpot force FR is proportional to the velocity v=dx/dt of the mass and acts opposite to the direction of motion. Then

FR=cv=cdxdt(with c>0),

so FR<0 if v>0 (motion to the right) while FR>0 if v<0 (motion to the left).

If FR and FS are the only forces acting on the mass m (Fig. 5.1.2) and its resulting acceleration is a=dv/dt, then Newton’s law F=ma gives

(4)mx=FS+FR;

FIGURE 5.1.2.

Directions of the forces acting on m.

that is,

(5)md2xdt2+cdxdt+kx=0.

Thus we have a differential equation satisfied by the position function x(t) of the mass m. This homogeneous second-order linear equation governs the free vibrations of the mass; we will return to this problem in detail in Section 5.4.

If, in addition to FS and FR, the mass m is acted on by an external force F(t)—which must then be added to the right-hand side in Eq. (4)—the resulting equation is

(6)md2xdt2+cdxdt+kx=F(t).

This nonhomogeneous linear differential equation governs the forced vibrations of the mass under the influence of the external force F(t).

Homogeneous Second-Order Linear Equations

Consider the general second-order linear equation

(7)A(x)y+B(x)y+C(x)y=F(x),

where the coefficient functions A, B, C, and F are continuous on the open interval I. Here we assume in addition that A(x)0 at each point of I, so we can divide each term in Eq. (7)by A(x) and write it in the form

(8)y+p(x)y+q(x)y=f(x).

We will discuss first the associated homogeneous equation

(9)y+p(x)y+q(x)y=0.

A particularly useful property of this homogeneous linear equation is the fact that the sum of any two solutions of Eq. (9) is again a solution, as is any constant multiple of a solution. This is the central idea of the following theorem.

Remark

According to Theorem 1 in Section 4.2, Theorem 1 here implies that the set S of all solutions of the homogeneous linear second-order equation (9) is a subspace of the vector space of all functions on the real line R—and hence is itself a vector space. We therefore call S the solution space of the differential equation.

Example 1

We can see by inspection that

y1(x)=cosxandy2(x)=sinx

are two solutions of the equation

y+y=0.

Theorem 1 tells us that any linear combination of these solutions, such as

y(x)=3y1(x)2y2(x)=3cosx2sinx,

is also a solution. We will see later that, conversely, every solution of y+y=0 is a linear combination of these two particular solutions y1 and y2. Thus a general solution of y+y=0 is given by

y(x)=c1cosx+c2sinx.

It is important to understand that this single formula for the general solution encompasses a “twofold infinity” of particular solutions, because the two coefficients c1 and c2 can be selected independently. Figures 5.1.3 through 5.1.5 illustrate some of the possibilities, with either c1 or c2 set equal to zero, or with both nonzero.

FIGURE 5.1.3.

Solutions y(x)=c1cosx of y+y=0.

FIGURE 5.1.4.

Solutions y(x)=c2cosx of y+y=0.

FIGURE 5.1.5.

Solutions of y+y=0 with c1 and c2 both nonzero.

Earlier in this section we gave the linear equation mx+cx+kx=F(t) as a mathematical model of the motion of the mass shown in Fig. 5.1.1. Physical considerations suggest that the motion of the mass should be determined by its initial position and initial velocity. Hence, given any preassigned values of x(0) and x(0), Eq. (6) ought to have a unique solution satisfying these initial conditions. More generally, in order to be a “good” mathematical model of a deterministic physical situation, a differential equation must have unique solutions satisfying any appropriate initial conditions. The following existence and uniqueness theorem (proved in the Appendix) gives us this assurance for the general second-order equation.

Remark 1

Equation (8) and the conditions in (11) constitute a second-order linear initial value problem. Theorem 2 tells us that any such initial value problem has a unique solution on the whole interval I where the coefficient functions in (8) are continuous. Recall from Section 1.3 that a nonlinear differential equation generally has a unique solution on only a smaller interval.

Remark 2

Whereas a first-order differential equation dy/dx=F(x,y) generally admits only a single solution curve y=y(x) passing through a given initial point (a, b), Theorem 2 implies that the second-order equation in (8) has infinitely many solution curves passing through the point (a,b0)—namely, one for each (real number) value of the initial slope y(a)=b1. That is, instead of there being only one line through (a,b0) tangent to a solution curve, every nonvertical straight line through (a,b0) is tangent to some solution curve of Eq. (8). Figure 5.1.6 shows a number of solution curves of the equation y+3y+2y=0 all having the same initial value y(0)=1, while Fig. 5.1.7 shows a number of solution curves all having the same initial slope y(0)=1. The application at the end of this section suggests how to construct such families of solution curves for a given homogeneous second-order linear differential equation.

FIGURE 5.1.6.

Solutions of y+3y+2y=0 with the same initial value y(0)=1 but different initial slopes.

FIGURE 5.1.7.

Solutions of y+3y+2y=0 with the same initial slope y(0)=1 but different initial values.

Example 1

Continued We saw in the first part of Example 1 that y(x)=3cos x2sin x is a solution (on the entire real line) of y+y=0. It has the initial values y(0)=3, y(0)=2. Theorem 2 tells us that this is the only solution with these initial values. More generally, the solution

y(x)=b0cosx+b1sinx

satisfies the arbitrary initial conditions y(0)=b0, y(0)=b1; this illustrates the existence of such a solution, also as guaranteed by Theorem 2.

Example 1 suggests how, given a homogeneous second-order linear equation, we might actually find the solution y(x) whose existence is assured by Theorem 2. First, we find two “essentially different” solutions y1 and y2; second, we attempt to impose on the general solution

(12)y=c1y1+c2y2

the initial conditions y(a)=b0, y(a)=b1. That is, we attempt to solve the simultaneous equations

(13)c1y1(a)+c2y2(a)=bo,c1y1(a)+c2y2(a)=b1

for the coefficients c1 and c2.

Example 2

Verify that the functions

y1(x)=exandy2(x)=xex

are solutions of the differential equation

y2y+y=0,

and then find a solution satisfying the initial conditions y(0)=3, y(0)=1.

Solution

The verification is routine; we omit it. We impose the given initial conditions on the general solution

y(x)=c1ex+c2xex,

for which

y(x)=(c1+c2)ex+c2xex,

to obtain the simultaneous equations

y(0)=c1=3,y(0)=c1+c2=1.

The resulting solution is c1=3, c2=2. Hence the solution of the original initial value problem is

y(x)=3ex2xex.

Figure 5.1.8 shows several additional solutions of y2y+y=0, all having the same initial value y(0)=3.

FIGURE 5.1.8.

Different solutions y(x)=3ex+c2xex of y2y+y=0 with the same initial value y(0)=3.

Linearly Independent Solutions

In order for the procedure of Example 2 to succeed, the two solutions y1 and y2 must have the property that the equations in (13) can always be solved for c1 and c2, no matter what the initial values y(a)=b0 and y(a)=b1 might be. This is so, provided that the two functions y1 and y2 are linearly independent. According to the following definition, linear independence of functions defined on an interval I is analogous to linear independence of ordinary vectors.

Two functions are said to be linearly dependent on an open interval provided that they are not linearly independent there; that is, one of them is a constant multiple of the other. We can always determine whether two given functions f and g are linearly dependent on an interval I by noting at a glance whether either of the two quotients f/g or g/f is a constant-valued function on I.

Example 3

Thus it is clear that the following pairs of functions are linearly independent on the entire real line:

sinxandcosx;exande2x;exandxex;x+1andx2;xand|x|.

That is, neither sin x/cos x=tanx nor cos x/sin x=cotx is a constant-valued function; neither ex/e2x=e3x nor e2x/ex is a constant-valued function; and so forth. But the identically zero function f(x)0 and any other function g are linearly dependent on every interval, because 0·g(x)=0=f(x). Also, the functions

f(x)=sin2xandg(x)=sinx cosx

are linearly dependent on any interval because f(x)=2g(x) is the familiar trigonometric identity sin 2x=2sin xcos x.

General Solutions

But does the homogeneous equation y+py+qy=0 always have two linearly independent solutions? Theorem 2 says yes! We need only choose y1 and y2 so that

y1(a)=1,y1(a)=0andy2(a)=0,y2(a)=1.

It is then impossible that either y1=ky2 or y2=ky1 because k·01 for any constant k. Theorem 2 tells us that two such linearly independent solutions exist; actually finding them is a crucial matter that we will discuss briefly at the end of this section, and in greater detail beginning in Section 5.3.

We want to show, finally, that given any two linearly independent solutions y1 and y2 of the homogeneous equation

(9)y(x)+p(x)y(x)+q(x)y(x)=0,

every solution y of Eq. (9) can be expressed as a linear combination

(12)y=c1y1+c2y2

of y1 and y2. This means that the function in (12) is a general solution of Eq. (9)—it provides all possible solutions of the differential equation.

As suggested by the equations in (13), the determination of the constants c1 and c2 in (12) depends on a certain 2×2 determinant of values of y1, y2, and their derivatives. Given two functions f and g, the Wronskian of f and g is the determinant

W=|fgfg|=fgfg.

We write either W(f, g) or W(x), depending on whether we wish to emphasize the two functions or the point x at which the Wronskian is to be evaluated. For example,

W(cosx,sinx)=|cosxsinxsinxcosx|=cos2x+sin2x=1

and

W(ex,xex)=|exxexexex+xex|=e2x.

These are examples of linearly independent pairs of solutions of differential equations (see Examples 1 and 2). Note that in both cases the Wronskian is everywhere nonzero.

On the other hand, if the functions f and g are linearly dependent, with f=kg (for example), then

W(f,g)=|kggkgg|=kggkgg0.

Thus the Wronskian of two linearly dependent functions is identically zero. In Section 5.2 we will prove that, if the two functions y1 and y2 are solutions of a homogeneous second-order linear equation, then the strong converse stated in part (b) of Theorem 3 holds.

Thus, given two solutions of Eq. (9), there are just two possibilities: The Wronskian W is identically zero if the solutions are linearly dependent; the Wronskian is never zero if the solutions are linearly independent. The latter fact is what we need to show that y=c1y1+c2y2 is the general solution of Eq. (9) if y1 and y2 are linearly independent solutions.

Remark

In essence, Theorem 4 tells us that, when we have found two linearly independent solutions y1 and y2 of the homogeneous equation in (9), we have then found all of its solutions. Specifically, {y1,y2} is then a basis for the solution space of the differential equation. We therefore call the linear combination Y=c1y1+c2y2 a general solution of the differential equation.

Example 4

If y1(x)=e2x and y2(x)=e2x, then

y1=(2)(2)e2x=4e2x=4y1andy2=(2)(2)e2x=4e2x=4y2.

Therefore, y1 and y2 are linearly independent solutions of

(15)y4y=0.

But y3(x)=cosh 2x and y4(x)=sinh 2x are also solutions of Eq. (15), because

d2dx2(cosh2x)=ddx(2sinh2x)=4cosh2x

and, similarly, (sinh2x)=4sinh2x. It therefore follows from Theorem 4 that the functions cosh 2 x and sinh 2 x can be expressed as linear combinations of y1(x)=e2x and y2(x)=e2x. Of course, this is no surprise, because

cosh 2x=12e2x+12e2xandsinh 2x=12e2x12e2x

by the definitions of the hyperbolic cosine and hyperbolic sine. Thus the solution space of the differential equation y4y=0 has the two different bases {e2x,e2x} and {cosh 2x, sinh 2x}.

Remark

Because e2x, e2x and cosh x, sinh x are two different pairs of linearly independent solutions of the equation y4y=0 in (15), Theorem 4 implies that every particular solution Y(x) of this equation can be written both in the form

Y(x)=c1e2x+c2e2x

and in the form

Y(x)=acosh 2x+bsinh 2x.

Thus these two different linear combinations (with arbitrary constant coefficients) provide two different descriptions of the set of all solutions of the same differential equation y4y=0. Hence each of these two linear combinations is a general solution of the equation. Indeed, this is why it is accurate to refer to a specific such linear combination as “a general solution” rather than as “the general solution.”

Linear Second-Order Equations with Constant Coefficients

As an illustration of the general theory introduced in this section, we discuss the homogeneous second-order linear differential equation

(16)ay+by+cy=0

with constant coefficients a, b, and c. We first look for a single solution of Eq. (16) and begin with the observation that

(17)(erx)=rerxand(erx)=r2erx,

so any derivative of erx is a constant multiple of erx. Hence, if we substituted y=erx in Eq. (16), then each term would be a constant multiple of erx, with the constant coefficients dependent on r and the coefficients a, b, and c. This suggests that we try to find a value of r so that these multiples of erx will have sum zero. If we succeed, then y=erx will be a solution of Eq. (16).

For example, if we substitute y=erx in the equation

y5y+6y=0,

we obtain

r2erx5rerx+6erx=0.

Thus

(r25r+6)erx=0;(r2)(r3)erx=0.

Hence y=erx will be a solution if either r=2 or r=3. So, in searching for a single solution, we actually have found two solutions: y1(x)=e2x and y2(x)=e3x.

To carry out this procedure in the general case, we substitute y=erx in Eq. (16). With the aid of the equations in (17), we find the result to be

ar2erx+brerx+cerx=0.

Because erx is never zero, we conclude that y(x)=erx will satisfy the differential equation in (16) precisely when r is a root of the algebraic equation

(18)ar2+br+c=0.

This quadratic equation is called the characteristic equation of the homogeneous linear differential equation

(16)ay+by+cy=0.

If Eq. (18) has two distinct (unequal) roots r1 and r2, then the corresponding solutions y1(x)=er1x and y2(x)=er2x of (16) are linearly independent. (Why?) This gives the following result.

Example 5

Find the general solution of

2y7y+3y=0.

Solution

We can solve the characteristic equation

2r27r+3=0

by factoring:

(2r1)(r3)=0.

The roots r1=12 and r2=3 are real and distinct, so Theorem 5 yields the general solution

y(x)=c1ex/2+c2e3x.

Example 6

The differential equation y+2y=0 has characteristic equation

r2+2r=r(r+2)=0

with distinct real roots r1=0 and r2=2. Because e0x1, we get the general solution

y(x)=c1+c2e2x.

Figure 5.1.9 shows several different solution curves with c1=1, all appearing to approach the solution curve y(x)1 (with c2=0) as x+.

FIGURE 5.1.9.

Solutions y(x)=1+c2e2x of y+2y=0 with different values of c2.

Remark

Note that Theorem 5 changes a problem involving a differential equation into one involving only the solution of an algebraic equation.

If the characteristic equation in (18) has equal roots r1=r2, we get (at first) only the single solution y1(x)=er1x of Eq. (16). The problem in this case is to produce the “missing” second solution of the differential equation.

A double root r=r1 will occur precisely when the characteristic equation is a constant multiple of the equation

(rr1)2=r22r1r+r12=0.

Any differential equation with this characteristic equation is equivalent to

(20)y2r1y+r12y=0.

But it is easy to verify by direct substitution that y=xer1x is a second solution of Eq. (20). It is clear (but you should verify) that

y1(x)=er1xandy2(x)=xer1x

are linearly independent functions, so the general solution of the differential equation in (20) is

y(x)=c1er1x+c2xer1x.

Example 7

To solve the initial value problem

y+2y+y=0;y(0)=5,y(0)=3,

we note first that the characteristic equation

r2+2r+1=(r+1)2=0

has equal roots r1=r2=1. Hence the general solution provided by Theorem 6 is

y(x)=c1ex+c2xex.

Differentiation yields

y(x)=c1ex+c2exc2xex,

so the initial conditions yield the equations

y(0)=c1=5,y(0)=c1+c2=3,

which imply that c1=5 and c2=2. Thus the desired particular solution of the initial value problem is

y(x)=5ex+2xex.

This particular solution, together with several others of the form y(x)=c1ex+2xex, is illustrated in Fig. 5.1.10.

FIGURE 5.1.10.

Solutions y(x)=c1ex+2xex of y+2y+y=0 with different values of c1.

The characteristic equation in (18) may have either real or complex roots. The case of complex roots will be discussed in Section 5.3.

5.1 Problems

In Problems 1 through 16, a homogeneous second-order linear differential equation, two functions y1 and y2, and a pair of initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y=c1y1+c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x.

  1. yy=0; y1=ex, y2=ex; y(0)=0, y(0)=5

  2. y9y=0; y1=e3x, y2=e3x; y(0)=1, y(0)=15

  3. y+4y=0; y1=cos 2x, y2=sin 2x; y(0)=3, y(0)=8

  4. y+25y=0; y1=cos 5x, y2=sin 5x; y(0)=10, y(0)=10

  5. y3y+2y=0; y1=ex, y2=e2x; y(0)=1, y(0)=0

  6. y+y6y=0; y1=e2x, y2=e3x; y(0)=7, y(0)=1

  7. y+y=0; y1=1, y2=ex; y(0)=2, y(0)=8

  8. y3y=0; y1=1, y2=e3x; y(0)=4, y(0)=2

  9. y+2y+y=0; y1=ex, y2=xex; y(0)=2, y(0)=1

  10. y10y+25y=0; y1=e5x, y2=xe5x; y(0)=3, y(0)=13

  11. y2y+2y=0; y1=ex cos x, y2=ex sin x; y(0)=0, y(0)=5

  12. y+6y+13y=0; y1=e3x cos 2x, y2=e3x sin 2x; y(0)=2, y(0)=0

  13. x2y2xy+2y=0; y1=x, y2=x2; y(1)=3, y(1)=1

  14. x2y+2xy6y=0; y1=x2, y2=x3; y(2)=10, y(2)=15

  15. x2yxy+y=0; y1=x, y2=xln x; y(1)=7, y(1)=2

  16. x2y+xy+y=0; y1=cos(ln x), y2=sin(ln x); y(1)=2, y(1)=3

The following three problems illustrate the fact that the superposition principle does not generally hold for nonlinear equations.

  1. Show that y=1/x is a solution of y+y2=0, but that if c0 and c1, then y=c/x is not a solution.

  2. Show that y=x3 is a solution of yy=6x4, but that if c21, then y=cx3 is not a solution.

  3. Show that y11 and y2=x are solutions of yy+(y)2=0, but that their sum y=y1+y2 is not a solution.

Determine whether the pairs of functions in Problems 20 through 26 are linearly independent or linearly dependent on the real line.

  1. f(x)=π, g(x)=cos2 x+sin2 x

  2. f(x)=x3, g(x)=x2|x|

  3. f(x)=1+x, g(x)=1+|x|

  4. f(x)=xex, g(x)=|x|ex

  5. f(x)=sin2 x, g(x)=1cos 2x

  6. f(x)=ex sin x, g(x)=ex cos x

  7. f(x)=2cos x+3sin x, g(x)=3cos x2sin x

  8. Let yp be a particular solution of the nonhomogeneous equation y+py+qy=f(x) and let yc be a solution of its associated homogeneous equation. Show that y=yc+yp is a solution of the given nonhomogeneous equation.

  9. With yp=1 and yc=c1cos x+c2sin x in the notation of Problem 27, find a solution of y+y=1 satisfying the initial conditions y(0)=1=y(0).

Problems 29 through 32 explore the propeties of the Wronskian.

  1. Show that y1=x2 and y2=x3 are two different solutions of x2y4xy+6y=0, both satisfying the initial conditions y(0)=0=y(0). Explain why these facts do not contradict Theorem 2 (with respect to the guaranteed uniqueness).

  2. (a) Show that y1=x3 and y2=|x3| are linearly independent solutions on the real line of the equation x2y3xy+3y=0. (b) Verify that W(y1,y2) is identically zero. Why do these facts not contradict Theorem 3?

  3. Show that y1=sinx2 and y2=cosx2 are linearly independent functions, but that their Wronskian vanishes at x=0. Why does this imply that there is no differential equation of the form y+p(x)y+q(x)y=0, with both p and q continuous everywhere, having both y1 and y2 as solutions?

  4. Let y1 and y2 be two solutions of A(x)y+B(x)y+C(x)y=0 on an open interval I where A, B, and C are continuous and A(x) is never zero. (a) Let W=W(y1,y2). Show that

    A(x)dwdx=(y1)(Ay2)(y2)(Ay1).

    Then substitute for Ay2 and Ay2 from the original differential equation to show that

    A(x)dwdx=B(x)W(x).

    (b) Solve this first-order equation to deduce Abel’s formula

    W(x)=K exp (B(x)A(x)dx),

    where K is a constant. (c) Why does Abel’s formula imply that the Wronskian W(y1,y2) is either zero everywhere or nonzero everywhere (as stated in Theorem 3)?

Apply Theorems 5 and 6 to find general solutions of the differential equations given in Problems 33 through 42. Primes denote derivatives with respect to x.

  1. y3y+2y=0

  2. y+2y15y=0

  3. y+5y=0

  4. 2y+3y=0

  5. 2yyy=0

  6. 4y+8y+3y=0

  7. 4y+4y+y=0

  8. 9y12y+4y=0

  9. 6y7y20y=0

  10. 35yy12y=0

Each of Problems 43 through 48 gives a general solution y(x) of a homogeneous second-order differential equation ay+by+cy=0 with constant coefficients. Find such an equation.

  1. y(x)=c1+c2e10x

  2. y(x)=c1e10x+c2e10x

  3. y(x)=c1e10x+c2xe10x

  4. y(x)=c1e10x+c2e100x

  5. y(x)=c1+c2x

  6. y(x)=ex(c1ex2+c2ex2)

Problems 49 and 50 deal with the solution curves of y+3y+2y=0 shown in Figs. 5.1.6 and 5.1.7.

  1. Find the highest point on the solution curve with y(0)=1 and y(0)=6 in Fig. 5.1.6.

  2. Figure 5.1.7 suggests that the solution curves shown all meet at a common point in the third quadrant. Assuming that this is indeed the case, find the coordinates of that point.

  3. A second-order Euler equation is one of the form

    (22)ax2y+bxy+cy=0

    where a, b, c are constants. (a) Show that if x>0, then the substitution v=ln x transforms Eq. (22) into the constant-coefficient linear equation

    (23)ad2ydv2+(ba)dydv+cy=0

    with independent variable v. (b) If the roots r1 and r2 of the characteristic equation of Eq. (23) are real and distinct, conclude that a general solution of the Euler equation in (22) is y(x)=c1xr1+c2xr2.

Make the substitution v=ln x of Problem 51 to find general solutions (for x>0) of the Euler equations in Problems 52–56.

  1. x2y+xyy=0

  2. x2y+2xy12y=0

  3. 4x2y+8xy3y=0

  4. x2y+xy=0

  5. x2y3xy+4y=0

5.1 Application Plotting Second-Order Solution Families

This application deals with the plotting by computer of families of solutions such as those illustrated in Figs. 5.1.6 and 5.1.7. Show first that the general solution of the differential equation

(1)y+3y+2y=0

is

(2)y(x)=c1ex+c2e2x.

This is equivalent to the graphing calculator result shown in Figure 5.1.11, and to the Wolfram|Alpha output generated by the simple query

y″ + 3y′ + 2y = 0

FIGURE 5.1.11.

TI-Nspire CX CAS screen showing the general solution of y+3y+2y=0.

Next show that the particular solution of Eq. (1) satisfying y(0)=a, y(0)=b corresponds to c1=2a+b and c2=ab, that is

(3)y(x)=(2a+b)ex(a+b)e2x.

For Fig. 5.1.6, we fix a=1, leading to the particular solution

(4)y(x)=(b+2)ex(b+1)e2x.

The Matlab loop

x = - 1 :  0.02 :  5 % x-vector from x = - 1 to x = 5
for b = -6 :  2 :  6 % for b = -6 to 6 with db = 2 do
   y = (b + 2)*exp(-x) - (b + 1)*exp(-2*x);
   plot(x,y)
end

was used to generate Fig. 5.1.6.

For Fig. 5.1.7, we instead fix b=1, leading to the particular solution

(5)y(x)=(2a+1)ex(a+1)e2x.

The Matlab loop

x = -2 :  0.02 :  4   % x-vector from x = -2 to x = 4
for a = -3 :  1 :  3  % for a = -3 to 3 with da = 1 do
   y = (2*a + 1)*exp(-x) - (a + 1)*exp(-2*x);
   plot(x,y)
end

was used to generate Fig. 5.1.7.

Computer systems, such as Maple and Mathematica, as well as graphing calculators, have commands to carry out for-loops such as the two shown here. Moreover, such systems often allow for interactive investigation, in which the solution curve is immediately redrawn in response to on-screen input. For example, Fig. 5.1.12 was generated using Matlab’s uicontrol command; moving the sliders allows the user to experiment with various combinations of the initial conditions y(0)=a and y(0)=b.

FIGURE 5.1.12.

Matlab graph of Eq. (3) with a=2 and b=1. Using the sliders, a and b can be changed interactively.

The Mathematica command

Manipulate[
    Plot[(2*a+b)*Exp[-x]+(-b-a)*Exp[-2*x],
    {x,-1,5}, PlotRange -> {-5,5}],
    {a,-3,3}, {b,-6,6}]

produces a similar display, as does the Maple command

Explore(plot((2*a+b)*exp(-x)+(-b-a)*exp(-2*x),
    x = -1..5, y=-5..5))

(after first bringing up a dialog box in which the ranges of values for a and b are specified). Likewise some graphing calculators feature a touchpad or other screen navigation method allowing the user to vary specified parameters in real time (see Fig. 5.1.13).

FIGURE 5.1.13.

TI-Nspire CX CAS graph of Eq. (3) with a=2 and b=1. The arrows allow a and b to be changed interactively.

Begin by either reproducing Figs. 5.1.6 and 5.1.7 or by creating an interactive display that shows the graph of (3) for any desired combination of a and b. Then, for each of the following differential equations, modify your commands to examine the family of solution curves satisfying y(0)=1, as well as the family of solution curves satisfying the initial condition y(0)=1.

  1. yy=0

  2. y3y+2y=0

  3. 2y+3y+y=0

  4. y+y=0 (see Example 1)

  5. y+2y+2y=0, which has general solution y(x)=ex(c1cos x+c2sin x)

5.2 General Solutions of Linear Equations

We now show that our discussion in Section 5.1 of second-order linear equations generalizes in a very natural way to the general nth-order linear differential equation of the form

(1)P0(x)y(n)+P1(x)y(n1)++Pn1(x)y+Pn(x)y=F(x).

Unless otherwise noted, we will always assume that the coefficient functions Pi(x) and F(x)are continuous on some open interval I (perhaps unbounded) where we wish to solve the equation. Under the additional assumption that P0(x)0 at each point of I, we can divide each term in Eq. (1) by P0(x) to obtain an equation with leading coefficient 1, of the form

(2)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=f(x).

The homogeneous linear equation associated with Eq. (2) is

(3)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=0.

Just as in the second-order case, a homogeneous nth-order linear differential equation has the valuable property that any superposition, or linear combination, of solutions of the equation is again a solution. The proof of the following theorem is essentially the same—a routine verification—as that of Theorem 1 of Section 5.1.

Example 1

It is easy to verify that the three functions

y1(x)=e3x,y2(x)=cos 2x,andy3(x)=sin 2x

are all solutions of the homogeneous third-order equation

y(3)+3y+4y+12y=0

on the entire real line. Theorem 1 tells us that any linear combination of these solutions, such as

y(x)=3y1(x)+3y2(x)2y3(x)=3e3x+3 cos 2x2 sin 2x,

is also a solution on the entire real line. We will see that, conversely, every solution of the differential equation of this example is a linear combination of the three particular solutions y1, y2, and y3. Thus a general solution is given by

y(x)=c1e3x+c2cos 2x+c3sin 2x.

Existence and Uniqueness of Solutions

We saw in Section 5.1 that a particular solution of a second-order linear differential equation is determined by two initial conditions. Similarly, a particular solution of an nth-order linear differential equation is determined by n initial conditions. The following theorem, proved in the Appendix, is the natural generalization of Theorem 2 of Section 5.1.

Equation (2) and the conditions in (5) constitute an nth-order initial value problem. Theorem 2 tells us that any such initial value problem has a unique solution on the whole interval I where the coefficient functions in (2) are continuous. It tells us nothing, however, about how to find this solution. In Section 5.3 we will see how to construct explicit solutions of initial value problems in the constant-coefficient case that occurs often in applications.

Example 1

Continued We saw earlier that

y(x)=3e3x+3 cos 2x2 sin 2x

is a solution of

y(3)+3y+4y+12y=0

on the entire real line. This particular solution has initial values y(0)=0, y(0)=5, and y(0)=39, and Theorem 2 implies that there is no other solution with these same initial values. Note that its graph (in Fig. 5.2.1) looks periodic on the right. Indeed, because of the negative exponent, we see that y(x)3 cos 2x2 sin 2x for large positive x.

FIGURE 5.2.1.

The particular solution y(x)=3e3x+3 cos 2x2 sin 2x.

Remark

Because its general solution involves the three arbitrary constants c1, c2, and c3, the third-order equation in Example 1 has a “threefold infinity” of solutions, including three families of especially simple solutions:

  • y(x)=c1e3x (obtained from the general solution with c2=c3=0),

  • y(x)=c2cos 2x (with c1=c3=0), and

  • y(x)=c3sin 2x (with c1=c2=0).

Alternatively, Theorem 2 suggests a threefold infinity of particular solutions corresponding to independent choices of the three initial values y(0)=b0, y(0)=b1, and y(0)=b2. Figures 5.2.2 through 5.2.4 illustrate three corresponding families of solutions—for each of which, two of these three initial values are zero.

FIGURE 5.2.2.

Solutions of y(3)+3y+4y+12y=0 with y(0)=y(0)=0 but with different values for y(0).

FIGURE 5.2.3.

Solutions of y(3)+3y+4y+12y=0 with y(0)=y(0)=0 but with different values for y(0).

FIGURE 5.2.4.

Solutions of y(3)+3y+4y+12y=0 with y(0)=y(0)=0 but with different values for y(0).

Note that Theorem 2 implies that the trivial solution y(x)0 is the only solution of the homogeneous equation

(3)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=0

that satisfies the trivial initial conditions

y(a)=y(a)==y(n1)(a)=0.

Example 2

It is easy to verify that

y1(x)=x2andy2(x)=x3

are two different solutions of

x2y4xy+6y=0,

and that both satisfy the initial conditions y(0)=y(0)=0. Why does this not contradict the uniqueness part of Theorem 2? It is because the leading coefficient in this differential equation vanishes at x=0, so this equation cannot be written in the form of Eq. (3) with coefficient functions continuous on an open interval containing the point x=0.

Linearly Independent Solutions

On the basis of our knowledge of general solutions of second-order linear equations, we anticipate that a general solution of the homogeneous nth-order linear equation

(3)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=0

will be a linear combination

(4)y=c1y1+c2y2++cnyn,

where y1, y2, , yn are particular solutions of Eq. (3). But these n particular solutions must be “sufficiently independent” that we can always choose the coefficients c1, c2, , cn in (4) to satisfy arbitrary initial conditions of the form in (5). The question is this: What should be meant by independence of three or more functions?

Recall that two functions f1 and f2 are linearly dependent if one is a constant multiple of the other; that is, if either f1=kf2 or f2=kf1 for some constant k. If we write these equations as

(1)f1+(k)f2=0or(k)f1+(1)f2=0,

we see that the linear dependence of f1 and f2 implies that there exist two constants c1 and c2 not both zero such that

(6)c1f1+c2f2=0.

Conversely, if c1 and c2 are not both zero, then Eq. (6) certainly implies that f1 and f2 are linearly dependent.

In analogy with Eq. (6), we say that n functions f1, f2, , fn are linearly dependent provided that some nontrivial linear combination

c1f1+c2f2++cnfn

of them vanishes identically; nontrivial means that not all of the coefficients c1, c2, , cn are zero (although some of them may be zero).

Thus linear dependence of functions on an interval I is precisely analogous to linear dependence of ordinary vectors (Section 4.3).

If not all the coefficients in Eq. (7) are zero, then clearly we can solve for at least one of the functions as a linear combination of the others, and conversely. Thus the functions f1, f2, , fn are linearly dependent if and only if at least one of them is a linear combination of the others.

Example 3

The functions

f1(x)=sin 2x,f2(x)=sin x cos x,andf3(x)=ex

are linearly dependent on the real line because

(1)f1+(2)f2+(0)f3=0

(by the familiar trigonometric identity sin 2x=2 sin x cos x).

The n functions f1, f2, , fn are called linearly independent on the interval I provided that they are not linearly dependent there. Equivalently, they are linearly independent on I provided that the identity

(7)c1f1+c2f2++cnfn=0

holds on I only in the trivial case

c1=c2==cn=0;

that is, no nontrivial linear combination of these functions vanishes on I. Put yet another way, the functions f1, f2, , fn are linearly independent if no one of them is a linear combination of the others. (Why?)

Sometimes one can show that n given functions are linearly dependent by finding, as in Example 3, nontrivial values of the coefficients so that Eq. (7) holds. But in order to show that n given functions are linearly independent, we must prove that nontrivial values of the coefficients cannot be found, and this is seldom easy to do in any direct or obvious manner.

Fortunately, in the case of n solutions of a homogeneous nth-order linear equation, there is a tool that makes the determination of their linear dependence or independence a routine matter in many examples. This tool is the Wronskian determinant, which we introduced (for the case n=2) in Section 5.1. Suppose that the n functions f1, f2, , fn are each n1 times differentiable. Then their Wronskian is the n×n determinant

(8)W=|f1f2fnf1f2fnf1(n1)f2(n1)fn(n1)|.

We write W(f1,f2,,fn) or W(x), depending on whether we wish to emphasize the functions or the point x at which their Wronskian is to be evaluated. The Wronskian is named after the Polish mathematician J. M. H. Wronski (1778–1853).

We saw in Section 5.1 that the Wronskian of two linearly dependent functions vanishes identically. More generally, the Wronskian of n linearly dependent functions f1, f2, , fn is identically zero. To prove this, assume that Eq. (7) holds on the interval I for some choice of the constants c1, c2, , cn not all zero. We then differentiate this equation n1 times in succession, obtaining the n equations

(9)c1f1(x)+c2f2(x)++cnfn(x)=0,c1f1(x)+c2f2(x)++cnfn(x)=0,c1f1(n1)(x)+c2f2(n1)(x)++cnfn(n1)(x)=0,

which hold for all x in I. We recall from Theorem 7 in Section 3.5 that a homogeneous n×n linear system of equations has a nontrivial solution if and only if its coefficient matrix is not invertible, which by Theorem 2 in Section 3.6 is so if and only if the coefficient determinant vanishes. In Eq. (9) the unknowns are the constants c1, c2, , cn and the determinant of coefficients is simply the Wronskian W(f1,f2,,fn) evaluated at the typical point x of I. Because we know that the ci are not all zero, it follows that W(x)0, as we wanted to prove.

Therefore, to show that the functions f1, f2, , fn are linearly independent on the interval I, it suffices to show that their Wronskian is nonzero at just one point of I.

Example 4

Show that the functions y1(x)=e3x, y2(x)=cos 2x, and y3(x)=sin 2x (of Example 1) are linearly independent.

Solution

Their Wronskian is

W=|e3xcos 2xsin 2x3e3x2 sin 2x2 cos 2x9e3x4 cos 2x4 sin 2x|=e3x|2 sin 2x2 cos 2x4 cos 2x4 sin 2x|+3e3x|cos 2xsin 2x4 cos 2x4 sin 2x|+9e3x|cos 2xsin 2x2 sin 2x2 cos 2x|=26e3x0.

Because W0 everywhere, it follows that y1, y2, and y3 are linearly independent on any open interval (including the entire real line).

Example 5

Show first that the three solutions

y1(x)=x,y2(x)=xln x,andy3(x)=x2

of the third-order equation

(10)x3y(3)x2y+2xy2y=0

are linearly independent on the open interval x>0. Then find a particular solution of Eq. (10) that satisfies the initial conditions

(11)y(1)=3,y(1)=2,y(1)=1.

Solution

Note that for x>0, we could divide each term in (10) by x3 to obtain a homogeneous linear equation of the standard form in (3). When we compute the Wronskian of the three given solutions, we find that

W=|xxln xx211+ln x2x01x2|=x.

Thus W(x)0 for x>0, so y1, y2, and y3 are linearly independent on the interval x>0. To find the desired particular solution, we impose the initial conditions in (11) on

y(x)=c1x+c2xln x+c3x2,y(x)=c1+c2(1+ln x)+2c3x,y(x)=0+c2x+2c3.

This yields the simultaneous equations

y(1)=c1+c3=3,y(1)=c1+c2+2c3=2,y(1)=c2+2c3=1;

we solve to find c1=1, c2=3, and c3=2. Thus the particular solution in question is

y(x)=x3xln x+2x2.

Provided that W(y1,y2,,yn)0, it turns out (Theorem 4) that we can always find values of the coefficients in the linear combination

y=c1y1+c2y2++cnyn

that satisfy any given initial conditions of the form in (5). Theorem 3 provides the necessary nonvanishing of W in the case of linearly independent solutions.

Remark

According to Problem 35, the Wronskian of Theorem 3 satisfies the first-order equation W=p1(x)W. Solution of this equation yields Abel’s formula

W(x)=Kexp(p1(x)dx)

for the homogeneous linear equation in (3). According as the constant K=0 or K0, it follows immediately from Abel’s formula that either W vanishes everywhere or W is nonzero everywhere.

General Solutions

We can now show that, given any fixed set of n linearly independent solutions of a homogeneous nth-order equation, every (other) solution of the equation can be expressed as a linear combination of those n particular solutions. Using the fact from Theorem 3 that the Wronskian of n linearly independent solutions is nonzero, the proof of the following theorem is essentially the same as the proof of Theorem 4 of Section 5.1 (the case n=2).

Remark

Theorem 4 tells us that, once we have found n linearly independent solutions y1,y2,,yn of the nth-order homogeneous linear equation in (3), we really have found all of its solutions. For then the solution space S of the equation is an n-dimensional vector space with basis {y1,y2,,yn}. Because every solution of (3) can be expressed as a linear combination of the form

(14)y=c1y1+c2y2++cnyn,

we call such a linear combination of n linearly independent particular solutions a general solution of the homogeneous linear differential equation.

Example 6

According to Example 4, the particular solutions y1(x)=e3x, y2(x)=cos 2x, and y3(x)=sin 2x of the linear differential equation y(3)+3y+4y+12y=0 are linearly independent. Now Theorem 2 says that—given b0, b1, and b2—there exists a particular solution y(x) satisfying the initial conditions y(0)=b0, y(0)=b1, and y(0)=b2. Hence Theorem 4 implies that this particular solution is a linear combination of y1, y2, and y3. That is, there exist coefficients c1, c2, and c3 such that

y(x)=c1e3x+c2cos 2x+c3sin 2x.

Upon successive differentiation and substitution of x=0, we discover that to find these coefficients, we need only solve the three linear equations

c1+c2=b0,3c1+2c3=b1,9c14c2=b2,

(See the application for this section.)

Nonhomogeneous Equations

We now consider the nonhomogeneous nth-order linear differential equation

(2)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=f(x)

with associated homogeneous equation

(3)y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=0.

Suppose that a single fixed particular solution yp of the nonhomogeneous equation in (2) is known, and that Y is any other solution of Eq. (2). If yc=Yyp, then subsitution of yc in the differential equation gives (using the linearity of differentiation)

yc(n)+p1yc(n1)++pn1yc+pnyc=[(Y(n)+p1Y(n1)++pn1Y+pnY][(yp(n)+p1yp(n1)++pn1yp+pnyp]=f(x)f(x)=0.

Thus yc=Yyp is a solution of the associated homogeneous equation in (3). Then

(14)Y=yc+yp,

and it follows from Theorem 4 that

(15)yc=c1y1+c2y2++cnyn,

where y1, y2, , yn are linearly independent solutions of the associated homogeneous equation. We call yc a complementary function of the nonhomogeneous equation and have thus proved that a general solution of the nonhomogeneous equation in (2) is the sum of its complementary function yc and a single particular solution yp of Eq. (2).

Example 7

It is evident that yp=3x is a particular solution of the equation

(17)y+4y=12x,

and that yc(x)=c1cos 2x+c2sin 2x is its complementary solution. Find a solution of Eq. (17) that satisfies the initial conditions y(0)=5, y(0)=7.

Solution

The general solution of Eq. (17) is

y(x)=c1cos 2x+c2sin 2x+3x.

Now

y(x)=2c1sin 2x+2c2cos 2x+3.

Hence the initial conditions give

y(0)=c1=5,y(0)=2c2+3=7.

We find that c1=5 and c2=2. Thus the desired solution is

y(x)=5 cos 2x+2 sin 2x+3x.

5.2 Problems

In Problems 1 through 6, show directly that the given functions are linearly dependent on the real line. That is, find a nontrivial linear combination of the given functions that vanishes identically.

  1. f(x)=2x, g(x)=3x2, h(x)=5x8x2

  2. f(x)=5, g(x)=23x2, h(x)=10+15x2

  3. f(x)=0, g(x)=sin x, h(x)=ex

  4. f(x)=17, g(x)=2 sin2 x, h(x)=3 cos2 x

  5. f(x)=17, g(x)=cos2 x, h(x)=cos 2x

  6. f(x)=ex, g(x)=cosh x, h(x)=sinh x

In Problems 7 through 12, use the Wronskian to prove that the given functions are linearly independent on the indicated interval.

  1. f(x)=1, g(x)=x, h(x)=x2; the real line

  2. f(x)=ex, g(x)=e2x, h(x)=e3x; the real line

  3. f(x)=ex, g(x)=cos x, h(x)=sin x; the real line

  4. f(x)=ex, g(x)=x2, h(x)=x2 ln x; x>0

  5. f(x)=x, g(x)=xex, h(x)=x2ex; the real line

  6. f(x)=x, g(x)=cos(ln x), h(x)=sin(ln x); x>0

In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.

  1. y(3)+2yy2y=0; y(0)=1, y(0)=2, y(0)=0; y1=ex, y2=ex, y3=e2x

  2. y(3)6y+11y6y=0; y(0)=0, y(0)=0, y(0)=3; y1=ex, y2=e2x, y3=e3x

  3. y(3)3y+3yy=0; y(0)=2, y(0)=0, y(0)=0; y1=ex, y2=xex, y3=x2ex

  4. y(3)5y+8y4y=0; y(0)=1, y(0)=4, y(0)=0; y1=ex, y2=e2x, y3=xe2x

  5. y(3)+9y=0; y(0)=3, y(0)=1, y(0)=2; y1=1, y2=cos 3x, y3=sin 3x

  6. y(3)3y+4y2y=0; y(0)=1, y(0)=0, y(0)=0; y1=ex, y2=ex cos x, y3=ex sin x.

  7. x3y(3)3x2y+6xy6y=0; y(1)=6, y(1)=14, y(1)=22; y1=x, y2=x2, y3=x3

  8. x3y(3)+6x2y+4xy4y=0; y(1)=1, y(1)=5, y(1)=11; y1=x, y2=x2, y3=x2 ln x

In Problems 21 through 24, a nonhomogeneous differential equation, a complementary solution yc, and a particular solution yp are given. Find a solution satisfying the given initial conditions.

  1. y+y=3x; y(0)=2, y(0)=2; yc=c1cos x+c2sin x; yp=3x

  2. y4y=12; y(0)=0, y(0)=10; yc=c1e2x+c2e2x; yp=3

  3. y2y3y=6; y(0)=3, y(0)=11; yc=c1ex+c2e3x; yp=2

  4. y2y+2y=2x; y(0)=4, y(0)=8; yc=c1ex cos x+c2ex sin x; yp=x+1

  5. Let Ly=y+py+qy. Suppose that y1 and y2 are two functions such that

    Ly1=f(x)andLy2=g(x).

    Show that their sum y=y1+y2 satisfies the nonhomogeneous equation Ly=f(x)+g(x).

  6. (a) Find by inspection particular solutions of the two nonhomogeneous equations

    y+2y=4andy+2y=6x.

    (b) Use the method of Problem 25 to find a particular solution of the differential equation y+2y=6x+4.

  7. Prove directly that the functions

    f1(x)1,f2(x)=x,andf3(x)=x2

    are linearly independent on the whole real line. (Suggestion: Assume that c1+c2x+c3x2=0. Differentiate this equation twice, and conclude from the equations you get that c1=c2=c3=0.)

  8. Generalize the method of Problem 27 to prove directly that the functions

    f0(x)1,f1(x)=x,f2(x)=x2,,fn(x)=xn

    are linearly independent on the real line.

  9. Use the result of Problem 28 and the definition of linear independence to prove directly that, for any constant r, the functions

    f0(x)=erx,f1(x)=xerx,,fn(x)=xnerx

    are linearly independent on the whole real line.

  10. Verify that y1=x and y2=x2 are linearly independent solutions on the entire real line of the equation

    x2y2xy+2y=0,

    but that W(x,x2) vanishes at x=0. Why do these observations not contradict part (b) of Theorem 3?

  11. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation

    y+py+qy=0,

    explain why the value of y(a) is determined by the values of y(a) and y(a). (b) Prove that the equation

    y2y5y=0

    has a solution satisfying the conditions

    y(0)=1,y(0)=0andy(0)=c

    if and only if C=5.

  12. Prove that an nth-order homogeneous linear differential equation satisfying the hypotheses of Theorem 2 has n linearly independent solutions y1,y2,,yn. (Suggestion: Let yi be the unique solution such that

    yi(i1)(a)=1andyi(k)(a)=0if ki1.)
  13. Suppose that the three numbers r1, r2, and r3 are distinct. Show that the three functions exp(r1x), exp(r2x), and exp(r3x) are linearly independent by showing that their Wronskian

    W=exp[(r1+r2+r3)x]|111r1r2r3r12r22r33|

    is nonzero for all x.

  14. Assume as known that the Vandermonde determinant

    V=|111r1r2rnr12r22rn2r1n1r2n1rnn1|

    is nonzero if the numbers r1,r2,,rn are distinct. Prove by the method of Problem 33 that the functions

    fi(x)=exp(rix),1in

    are linearly independent.

  15. According to Problem 32 of Section 5.1, the Wronskian W(y1,y2) of two solutions of the second-order equation

    y+p1(x)y+p2(x)y=0

    is given by Abel’s formula

    W(x)=Kexp(p1(x)dx)

    for some constant K. It can be shown that the Wronskian of n solutions y1,y2,,yn of the nth-order equation

    y(n)+p1(x)y(n1)++pn1(x)y+pn(x)y=0

    satisfies the same identity. Prove this for the case n=3 as follows: (a) The derivative of a determinant of functions is the sum of the determinants obtained by separately differentiating the rows of the original determinant. Conclude that

    W=|y1y2y3y1y2y3y1(3)y2(3)y3(3)|.

    (b) Substitute for y1(3), y2(3), and y3(3) from the equation

    y(3)+p1y+p2y+p3y=0,

    and then show that W=p1W. Integration now gives Abel’s formula.

  16. Suppose that one solution y1(x) of the homogeneous second-order linear differential equation

    (18)y+p(x)y+q(x)y=0

    is known (on an interval I where p and q are continuous functions). The method of reduction of order consists of substituting y2(x)=v(x)y1(x) in (18) and attempting to determine the function v(x) so that y2(x) is a second linearly independent solution of (18). After substituting y=v(x)y1(x) in Eq. (18), use the fact that y1(x) is a solution to deduce that

    (19)y1v+(2y1+py1)v=0.

    If y1(x) is known, then (19) is a separable equation that is readily solved for the derivative v(x) of v(x). Integration of v(x) then gives the desired (nonconstant) function v(x).

  17. Before applying Eq. (19) with a given homogeneous second-order linear differential equation and a known solution y1(x), the equation must first be written in the form of (18) with leading coefficient 1 in order to correctly determine the coefficient function p(x). Frequently it is more convenient to simply substitute y=v(x)y1(x) in the given differential equation and then proceed directly to find v(x). Thus, starting with the readily verified solution y1(x)=x3 of the equation

    x2y5xy+9y=0(x>0),

    substitute y=vx3 and deduce that xv+v=0. Thence solve for v(x)=Cln x, and thereby obtain (with C=1) the second solution y2(x)=x3 ln x.

In each of Problems 38 through 42, a differential equation and one solution y1 are given. Use the method of reduction of order as in Problem 37 to find a second linearly independent solution y2.

  1. x2y+xy9y=0 (x>0)y1(x)=x3

  2. 4y4y+y=0y1(x)=ex/2

  3. x2yx(x+2)y+(x+2)y=0 (x>0)y1(x)=x

  4. (x+1)y(x+2)y+y=0 (x>1)y1(x)=ex

  5. (1x2)y+2xy2y=0 (1<x<1)y1(x)=x

  6. First note that y1(x)=x is one solution of Legendre’s equation of order 1,

    (1x2)y2xy+2y=0.

    Then use the method of reduction of order to derive the second solution

    y2(x)=1x2ln1+x1x(for1<x<1).
  7. First verify by substitution that y1(x)=x1/2 cos x is one solution (for x>0) of Bessel’s equation of order 12,

    x2y+xy+(x214)y=0.

    Then derive by reduction of order the second solution y2(x)=x1/2 sin x.

5.2 Application Plotting Third-Order Solution Families

This application deals with the plotting by computer of families of solutions such as those illustrated in Figs. 5.2.2 through 5.2.4. We know from Example 6 that the general solution of

(1)y(3)+3y+4y+12y=0

is

(2)y(x)=c1e3x+c2 cos 2x+c3 sin 2x.

For Fig. 5.2.2, use the method of Example 6 to show that the particular solution of Eq. (1)satisfying the initial conditions y(0)=a, y(0)=0, and y(0)=0 is given by

(3)y(x)=a13(4e3x+9 cos 2x+6 sin 2x).

The Matlab loop

x = -1.5 :  0.02 :  5 % x-vector from x = -1.5 to x = 5
for a = -3 :  1 :  3  % for a = -3 to 3 with da = 1 do
   c1 = 4*a/13;
   c2 = 9*a/13;
   c3 = 6*a/13;
   y = c1*exp(-3*x) + c2*cos(2*x) + c3*sin(2*x);
   plot(x,y)
end

was used to generate Fig. 5.2.2.

For Fig. 5.2.3, show that the particular solution of Eq. (1) satisfying the initial conditions y(0)=0, y(0)=b, and y(0)=0 is given by

(4)y(x)=b2 sin 2x,

and alter the preceding for-loop accordingly.

For Fig. 5.2.4, show that the particular solution of Eq. (1) satisfying the initial conditions y(0)=0, y(0)=0, and y(0)=c is given by

(5)y(x)=c26(2e3x2 cos 2x+3 sin 2x).

Computer algebra systems such as Maple and Mathematica, as well as graphing calculators, have commands to carry out for-loops such as the one shown here. Begin by reproducing Figs. 5.2.2 through 5.2.4. Then plot similar families of solution curves for the differential equations in Problems 13 through 20.

5.3 Homogeneous Equations with Constant Coefficients

In Section 5.2 we saw that a general solution of an nth-order homogeneous linear equation is a linear combination of n linearly independent particular solutions, but we said little about how actually to find even a single solution. The solution of a linear differential equation with variable coefficients ordinarily requires numerical methods (Chapter 2) or infinite series methods (Chapter 8). But we can now show how to find, explicitly and in a rather straightforward way, n linearly independent solutions of a given nth-order linear equation if it has constant coefficients. The general such equation may be written in the form

(1)any(n)+an1y(n1)++a2y+a1y+a0y=0,

where the coefficients a0, a1, a2, , an are real constants with an0.

The Characteristic Equation

We first look for a single solution of Eq. (1), and begin with the observation that

(2)dkdxk(erx)=rkerx,

so any derivative of erx is a constant multiple of erx. Hence, if we substituted y=erx in Eq. (1), each term would be a constant multiple of erx, with the constant coefficients depending on r and the coefficients ak. This suggests that we try to find r so that all these multiples of erx will have sum zero, in which case y=erx will be a solution of Eq. (1).

For example, in Section 5.1 we substituted y=erx in the second-order equation

ay+by+cy=0

to derive the characteristic equation

ar2+br+c=0

that r must satisfy.

To carry out this technique in the general case, we substitute y=erx in Eq. (1), and with the aid of Eq. (2) we find the result to be

anrnerx+an1rn1erx++a2r2erx+a1rerx+a0erx=0;

that is,

erx(anrn+an1rn1++a2r2+a1r+a0)=0.

Because erx is never zero, we see that y=erx will be a solution of Eq. (1) precisely when r is a root of the equation

(3)anrn+an1rn1++a2r2+a1r+a0=0.

This equation is called the characteristic equation or auxiliary equation of the differential equation in (1). Our problem, then, is reduced to the solution of this purely algebraic equation.

According to the fundamental theorem of algebra, every nth-degree polynomial—such as the one in Eq. (3)—has n zeros, though not necessarily distinct and not necessarily real. Finding the exact values of these zeros may be difficult or even impossible; the quadratic formula is sufficient for second-degree equations, but for equations of higher degree we may need either to spot a fortuitous factorization or to apply a numerical technique such as Newton’s method (or use a calculator/computer solve command).

Distinct Real Roots

Whatever the method we use, let us suppose that we have solved the characteristic equation. Then we can always write a general solution of the differential equation. The situation is slightly more complicated in the case of repeated roots or complex roots of Eq. (3), so let us first examine the simplest case—in which the characteristic equation has n distinct (no two equal) real roots r1, r2, , rn. Then the functions

er1x,er2x,,ernx

are all solutions of Eq. (1), and (by Problem 34 of Section 5.2) these n solutions are linearly independent on the entire real line. In summary, we have proved Theorem 1.

Example 1

Solve the initial value problem

y(3)+3y10y=0;y(0)=7,y(0)=0,y(0)=70.

Solution

The characteristic equation of the given differential equation is

r3+3r210r=0.

We solve by factoring:

r(r2+3r10)=r(r+5)(r2)=0,

and so the characteristic equation has the three distinct real roots r=0, r=5, and r=2. Because e0=1, Theorem 1 gives the general solution

y(x)=c1+c2e5x+c3e2x.

Then the given initial conditions yield the linear equations

y(0)=c1+c2+c3=7,y(0)=5c2+2c3=0,y(0)=25c2+4c3=70

in the coefficients c1, c2, and c3. The last two equations give y(0)2y(0)=35c2=70, so c2=2. Then the second equation gives c3=5, and finally the first equation gives c1=0. Thus the desired particular solution is

y(x)=2e5x+5e2x.

Polynomial Differential Operators

If the roots of the characteristic equation in (3) are not distinct—there are repeated roots—then we cannot produce n linearly independent solutions of Eq. (1) by the method of Theorem 1. For example, if the roots are 1, 2, 2, and 2, we obtain only the two functions ex and e2x. The problem, then, is to produce the missing linearly independent solutions. For this purpose, it is convenient to adopt “operator notation” and write Eq. (1) in the form Ly=0, where the operator

(5)L=andndxn+an1dn1dxn1++a2d2dx2+a1ddx+a0

operates on the n-times differentiable function y(x) to produce the linear combination

Ly=any(n)+an1y(n1)++a2y(2)+a1y+a0y

of y and its first n derivatives. We also denote by D=d/dx the operation of differentiation with respect to x, so that

Dy=y, D2y=y, D3y=y(3),

and so on. In terms of D, the operator L in (5) may be written

(6)L=anDn+an1Dn1++a2D2+a1D+ao,

and we will find it useful to think of the right-hand side in Eq. (6) as a (formal) nth-degree polynomial in the “variable” D; it is a polynomial differential operator.

A first-degree polynomial operator with leading coefficient 1 has the form Da, where a is a real number. It operates on a function y=y(x) to produce

(Da)y=Dyay=yay.

The important fact about such operators is that any two of them commute:

(7)(Da)(Db)y=(Db)(Da)y

for any twice differentiable function y=y(x). The proof of the formula in (7) is the following computation:

(Da)(Db)y=(Da)(yby)=D(yby)a(yby)=y(b+a)y+aby=y(a+b)y+bay=D(yay)b(yay)=(Db)(yay)=(Db)(Da)y.

We see here also that (Da)(Db)=D2(a+b)D+ab. Similarly, it can be shown by induction on the number of factors that an operator product of the form (Da1)(Da2)(Dan) expands—by multiplying out and collecting coefficients—in the same way as does an ordinary product (xa1)(xa2)(xan) of linear factors, with x denoting a real variable. Consequently, the algebra of polynomial differential operators closely resembles the algebra of ordinary real polynomials.

Repeated Real Roots

Let us now consider the possibility that the characteristic equation

(3)anrn+an1rn1++a2r2+a1r+a0=0

has repeated roots. For example, suppose that Eq. (3) has only two distinct roots, r0 of multiplicity 1 and r1 of multiplicity k=n1>1. Then (after dividing by an) Eq. (3) can be rewritten in the form

(8)(rr1)k(rr0)=(rro)(rr1)k=0.

Similarly, the corresponding operator L in (6) can be written as

(9)L=(Dr1)k(Dr0)=(Dro)(Dr1)k,

the order of the factors making no difference because of the formula in (7).

Two solutions of the differential equation Ly=0 are certainly y0=er0x and y1=er1x. This is, however, not sufficient; we need k+1 linearly independent solutions in order to construct a general solution, because the equation is of order k+1. To find the missing k1 solutions, we note that

Ly=(Dr0)[(Dr1)ky]=0.

Consequently, every solution of the kth-order equation

(10)(Dr1)ky=0

will also be a solution of the original equation Ly=0. Hence our problem is reduced to that of finding a general solution of the differential equation in (10).

The fact that y1=er1x is one solution of Eq. (10) suggests that we try the substitution

(11)y(x)=u(x)y1(x)=u(x)er1x,

where u(x) is a function yet to be determined. Observe that

(12)(Dr1)[uer1x]=(Du)er1x+u(r1er1x)r1(uer1x)=(Du)er1x.

Upon k applications of this fact, it follows that

(13)(Dr1)k[uer1x]=(Dku)er1x

for any sufficiently differentiable function u(x). Hence y=uer1x will be a solution of Eq. (10) if and only if Dku=u(k)=0. But this is so if and only if

u(x)=c1+c2x+c3x2++ckxk1,

a polynomial of degree at most k1. Hence our desired solution of Eq. (10) is

y(x)=uer1x=(c1+c2x+c3x2++ckxk1)er1x.

In particular, we see here the additional solutions xer1x, x2er1x, , xk1er1x of the original differential equation Ly=0.

The preceding analysis can be carried out with the operator Dr1 replaced with an arbitrary polynomial operator. When this is done, the result is a proof of the following theorem.

Remark

According to Problem 29 of Section 5.2, the k functions erx, xerx, x2erx, , xk1erx involved in (14) are linearly independent. Thus a root of multiplicity k1 corresponds to k linearly independent “basis solutions” that generate a k-dimensional subspace of the n-dimensional solution space S of the nth-order homogeneous linear differential equation. Because the sum of the multiplicities of the separate roots of the nth-degree characteristic equation is n, we get a basis {y1,y2,,yn} for S when we assemble all the basis solutions corresponding to these separate roots. We can then write a general solution y=c1y1+c2y2++cnyn of the differential equation. In effect, we thereby construct a general solution by combining different “parts” corresponding to the separate roots of the characteristic equation.

Example 2

Find a general solution of the fifth-order differential equation

9y(5)6y(4)+y(3)=0.

Solution

The characteristic equation is

9r56r4+r3=r3(9r26r+1)=r3(3r1)2=0

It has the triple root r=0 and the double root r=13. The triple root r=0 contributes

c1e0.x+c2xe0.x+c3x2e0.x=c1+c2x+c3x2

to the solution, while the double root r=13 contributes c4ex/3+c5xex/3. Hence a general solution of the given differential equation is

y(x)=c1+c2x+c3x2+c4ex/3+c5xex/3.

Complex-Valued Functions and Euler’s Formula

Because we have assumed that the coefficients of the differential equation and its characteristic equation are real, any complex (nonreal) roots will occur in complex conjugate pairs a±bi where a and b are real and i=1. This raises the question of what might be meant by an exponential such as e(a+bi)x.

To answer this question, we recall from elementary calculus the Taylor (or MacLaurin) series for the exponential function

et=n=0tnn!=1+t+t22!+t33!+t44!+.

If we substitute t=iθ in this series and recall that i2=1, i3=i, i4=1, and so on, we get

eiθ=n=0(iθ)nn!=1+iθθ22!iθ33!+θ44!+iθ55!=(1θ22!+θ44!)+i(θθ33!+θ55!).

Because the two real series in the last line are the Taylor series for cos θ and sin θ, respectively, this implies that

(15)eiθ= cos  θ+i  sin  θ.

This result is known as Euler’s formula. Because of it, we define the exponential function ez, for z=x+iy an arbitrary complex number, to be

(16)ez=ex+iy=exeiy=ex(cos y+i sin y).

Thus it appears that complex roots of the characteristic equation will lead to complex-valued solutions of the differential equation. A complex-valued function F of the real variable x associates with each real number x (in its domain of definition) the complex number

(17)F(x)=f(x)+ig(x).

The real-valued functions f and g are called the real and imaginary parts, respectively, of F. If they are differentiable, we define the derivative F of F by

(18)F(x)=f(x)+ig(x).

Thus we simply differentiate the real and imaginary parts of F separately.

We say that the complex-valued function F(x) satisfies the homogeneous linear differential equation L[F(x)]=0 provided that its real and imaginary parts in (17) separately satisfy this equation—so L[F(x)]=L[f(x)]+iL[g(x)]=0.

The particular complex-valued functions of interest here are of the form F(x)=erx, where r=a±bi. We note from Euler’s formula that

(19a)e(a+bi)x=eaxeibx=eax(cos bx+i sin bx)

and

(19b)e(abi)x=eaxeibx=eax(cos bxi sin abx).

The most important property of erx is that

(20)Dx(erx)=rerx,

if r is a complex number. The proof of this assertion is a straightforward computation based on the definitions and formulas given earlier:

Dx(erx)=Dx(eax cos bx)+iDx(eax sin bx)=(aeax cosbxbeax sinbx)+i (aeax sinbx+beax cos bx)=(a+bi)(eax cos bx+i eax sin bx)=rerx.

Complex Roots

It follows from Eq. (20) that when r is complex (just as when r is real), erx will be a solution of the differential equation in (1) if and only if r is a root of its characteristic equation. If the complex conjugate pair of roots r1=a+bi and r2=abi are simple (nonrepeated), then the corresponding part of a general solution of Eq. (1) is

y(x)=C1er1x+C2er2x=C1e(a+bi)x+C2e(abi)x=C1eax(cos bx+i sin bx)+C2eax(cos bxi sin bx)y(x)=(C1+C2)eaxcos  bx+i (C1C2)eaxsin bx,

where the arbitrary constants C1 and C2 can be complex. For instance, the choice C1=C2=12 gives the real-valued solution y1(x)=eax cos bx, while the choice C1=12i, C2=12i gives the independent real-valued solution y2(x)=eax sin bx. This yields the following result.

Example 3

The characteristic equation of

y+b2y=0 (b>0)

is r2+b2=0, with roots r=±bi. So Theorem 3 (with a=0) gives the general solution

y(x)=c1 cos bx+c2 sin bx.

Example 4

Find the particular solution of

y4y+5y=0

for which y(0)=1 and y(0)=5.

Solution

Completion of the square in the characteristic equation yields

r24r+5=(r2)2+1=0,

so r2=±1=±i. Thus we obtain the complex conjugate roots 2±i (which could also be found directly using the quadratic formula). Hence Theorem 3 with a=2 and b=1 gives the general solution

y(x)=e2x(c1cos x+c2sin x).

Then

y(x)=2e2x(c1cos x+c2sin x)+e2x(c1sin x+c2cos x),

so the initial conditions give

y(0)=c1=1andy(0)=2c1+c2=5.

It follows that c2=3, and so the desired particular solution is

y(x)=e2x(cos x+3 sin x).

In Example 5 below we employ the polar form

(22)z=x+iy=reiθ

of the complex number z. This form follows from Euler’s formula upon writing

z=r(xr+iyr)=r(cos θ+i sin θ)=reiθ

in terms of the modulus r=x2+y2>0 of the number z and its argument θ indicated in Fig. 5.3.1. For instance, the imaginary number i has modulus 1 and argument π/2, so i=eiπ/2. Similarly, i=e3π/2. Another consequence is the fact that the nonzero complex number z=reiθ has the two square roots

(23)z=±(reiθ)1/2=±reiθ/2,

FIGURE 5.3.1.

Modulus and argument of the complex number x+iy.

where r denotes (as usual for a positive real number) the positive square root of the modulus of z.

Example 5

Find a general solution of y(4)+4y=0.

Solution

The characteristic equation is

r4+4=(r2)2(2i)2=(r2+2i)(r22i)=0,

and its four roots are ±±2i. Since i=eiπ/2 and i=ei3π/2, we find that

2i=(2eiπ/2)1/2=2eiπ/4=2(cosπ4+isinπ4)=1+i

and

2i=(2ei3π/2)1/2=2ei3π/4=2(cos3π4+isin3π4)=1+i.

Thus the four (distinct) roots of the characteristic equation are r=±(1+i) and r=±(1+i). Grouping these four into the complex conjugate pairs 1±i and 1±i yields the general solution

y(x)=ex(c1cosx+c2sinx)+ex(c3cosx+c4sinx)

of the differential equation y(4)+4y=0.

Repeated Complex Roots

Theorem 2 holds for repeated complex roots. If the conjugate pair a±bi has multiplicity k, then the corresponding part of the general solution has the form

(24)(A1+A2x++Akxk1)e(a+bi)x+(B1+B2x++Bkxk1)e(abi)x=p=0k1xpeax(cpcosbx+dpsinbx).

It can be shown that the 2k functions

xpeaxcosbx,xpeaxsinbx,0pk1

that appear in Eq. (24) are linearly independent.

Example 6

Find a general solution of (D2+6D+13)2y=0.

Solution

By completing the square, we see that the characteristic equation

(r2+6r+13)2=[(r+3)2+4]2=0

has as its roots the conjugate pair 3±2i of multiplicity k=2. Hence Eq. (24) gives the general solution

y(x)=e3x(c1cos2x+d1sin2x)+xe3x(c2cos2x+d2sin2x).

In applications we are seldom presented in advance with a factorization as convenient as the one in Example 6. Often the most difficult part of solving a homogeneous linear equation is finding the roots of its characteristic equation. Example 7 illustrates an approach that may succeed when a root of the characteristic equation can be found by inspection. The application material for this section illustrates other possibilities.

Example 7

The characteristic equation of the differential equation

y(3)+y10y=0

is the cubic equation

r3+r10=0.

By a standard theorem of elementary algebra, the only possible rational roots are the factors ±1, ±2, ±5, and ±10 of the constant term 10. By trial and error (if not by inspection) we discover the root 2. The factor theorem of elementary algebra implies that r2 is a factor of r3+r10, and division of the former into the latter produces as quotient the quadratic polynomial

r2+2r+5=(r+1)2+4.

The roots of this quotient are the complex conjugates 1±2i. The three roots we have found now yield the general solution

y(x)=c1e2x+ex(c2cos2x+c3sin2x).

Example 8

The roots of the characteristic equation of a certain differential equation are 3,5, 0, 0, 0, 0,5, 2±3i, and 2±3i. Write a general solution of this homogeneous differential equation.

Solution

The solution can be read directly from the list of roots. It is

y(x)=c1+c2x+c3x2+c4x3+c5e3x+c6e5x+c7xe5x+e2x(c8cos3x+c9sin3x)+xe2x(c10cos3x+c11sin3x).

5.3 Problems

Find the general solutions of the differential equations in Problems 1 through 20.

  1. y4y=0

  2. 2y3y=0

  3. y+3y10y=0

  4. 2y7y+3y=0

  5. y+6y+9y=0

  6. y+5y+5y=0

  7. 4y12y+9y=0

  8. y6y+13y=0

  9. y+8y+25y=0

  10. 5y(4)+3y(3)=0

  11. y(4)8y(3)+16y=0

  12. y(4)3y(3)+3yy=0

  13. 9y(3)+12y+4y=0

  14. y(4)+3y4y=0

  15. y(4)8y+16y=0

  16. y(4)+18y+81y=0

  17. 6y(4)+11y+4y=0

  18. y(4)=16y

  19. y(3)+yyy=0

  20. y(4)+2y(3)+3y+2y+y=0 (Suggestion: Expand (r2+r+1)2.)

Solve the initial value problems given in Problems 21 through 26.

  1. y4y+3y=0; y(0)=7, y(0)=11

  2. 9y+6y+4y=0; y(0)=3, y(0)=4

  3. y6y+25y=0; y(0)=3, y(0)=1

  4. 2y(3)3y2y=0; y(0)=1, y(0)=1, y(0)=3

  5. 3y(3)+2y=0; y(0)=1, y(0)=0, y(0)=1

  6. y(3)+10y+25y=0; y(0)=3, y(0)=4, y(0)=5

Find general solutions of the equations in Problems 27 through 32. First find a small integral root of the characteristic equation by inspection; then factor by division.

  1. y(3)+3y4y=0

  2. 2y(3)y5y2y=0

  3. y(3)+27y=0

  4. y(4)y(3)+y3y6y=0

  5. y(3)+3y+4y8y=0

  6. y(4)+y(3)3y5y2y=0

In Problems 33 through 36, one solution of the differential equation is given. Find the general solution.

  1. y(3)+3y54y=0; y=e3x

  2. 3y(3)2y+12y8y=0; y=e2x/3

  3. 6y(4)+5y(3)+25y+20y+4y=0; y=cos 2x

  4. 9y(3)+11y+4y14y=0; y=ex sin x

  5. Find a function y(x) such that y(4)(x)=y(3)(x) for all x and y(0)=18, y(0)=12, y(0)=13, and y(3)(0)=7.

  6. Solve the initial value problem

    y(3)5y+100y500y=0;y(0)=0 ,y(0)=10,y(0)=250

    given that y1(x)=e5x is one particular solution of the differential equation.

In Problems 39 through 42, find a linear homogeneous constant-coefficient equation with the given general solution.

  1. y(x)=(A+Bx+Cx2)e2x

  2. y(x)=Ae2x+B cos 2x+C sin 2x

  3. y(x)=A cos 2x+B sin 2x+Ccosh 2x+Dsinh 2x

  4. y(x)=(A+Bx+Cx2) cos 2x+(D+Ex+Fx2) sin 2x

Problems 43 through 47 pertain to the solution of differential equations with complex coefficients.

  1. (a) Use Euler’s formula to show that every complex number can be written in the form reiθ, where r0 and π<θπ. (b) Express the numbers 4,2, 3i, 1+i, and 1+i3 in the form reiθ. (c) The two square roots of reiθ are ±reiθ/2. Find the square roots of the numbers 22i3 and 2+2i3.

  2. Use the quadratic formula to solve the following equations. Note in each case that the roots are not complex conjugates.

    1. x2+ix+2=0

    2. x22ix+3=0

  3. Find a general solution of y2iy+3y=0.

  4. Find a general solution of yiy+6y=0.

  5. Find a general solution of y=(2+2i3)y.

  6. Solve the initial value problem

    y(3)=y;y(0)=1,y(0)=y(0)=0.

    (Suggestion: Impose the given initial conditions on the general solution

    y(x)=Aex+Beαx+Ceβx,

    where α and β are the complex conjugate roots of r31=0, to discover that

    y(x)=13(ex+2ex/2cosx32)

    is a solution.)

  7. Solve the initial value problem

    y(4)=y(3)+y+y+2y;y(0)=y(0)=y(0)=0, 2y(3)(0)=30.
  8. The differential equation

    (25)y+(sgn x)y=0

    has the discontinuous coefficient function

    sgn x={+1if x>0,1if x<0.

    Show that Eq. (25) nevertheless has two linearly independent solutions y1(x) and y2(x) defined for all x such that

    • Each satisfies Eq. (25) at each point x0;

    • Each has a continuous derivative at x=0;

    • y1(0)=y2(0)=1 and y2(0)=y1(0)=0.

    (Suggestion: Each yi(x) will be defined by one formula for x<0 and by another for x0.) The graphs of these two solutions are shown in Fig. 5.3.2.

    FIGURE 5.3.2.

    Graphs of y1(x) and y2(x) in Problem 50.

  9. According to Problem 51 in Section 5.1, the substitution v=ln x (x>0) transforms the second-order Euler equation ax2y+bxy+cy=0 to a constant-coefficient homogeneous linear equation. Show similarly that this same substitution transforms the third-order Euler equation

    ax3y+bx2y+cxy+dy=0

    (where a, b, c, d are constants) into the constant-coefficient equation

    ad3ydv3+(b3a)d2ydv2+(cb+2a)dydv+dy=0.

Make the substitution v=ln x of Problem 51 to find general solutions (for x>0) of the Euler equations in Problems 52 through 58.

  1. x2y+xy+9y=0

  2. x2y+7xy+25y=0

  3. x3y+6x2y+4xy=0

  4. x3yx2y+xy=0

  5. x3y+3x2y+xy=0

  6. x3y3x2y+xy=0

  7. x3y+6x2y+7xy+y=0

5.3 Application Approximate Solutions of Linear Equations

To meet the needs of applications such as those of this section, polynomial-solving utilities are now a common feature of calculator and computer systems and can be used to solve a characteristic equation numerically even when no simple factorization is evident or even possible. For instance, suppose that we want to solve the homogeneous linear differential equation

(1)y(3)3y+y=0

with characteristic equation

(2)r33r2+1=0.

A typical graphing calculator has a solve command that can be used to find the approximate roots of a polynomial equation. As indicated in Figs. 5.3.3 and 5.3.4, we find that the roots of Eq. (2) are given by r0.5321, 0.6527, and 2.8794. Some analogous computer algebra system commands are

FIGURE 5.3.3.

Solving the equation r33r2+1=0 with a TI-84 Plus CE calculator that requires an estimate of each root.

FIGURE 5.3.4.

Solving the characteristic equation with a TI-89 calculator having a more sophisticated solve facility.


fsolve(r^3 - 3*r^2 + 1 = 0, r);         (Maple)
NSolve[r^3 - 3*r^2 + 1 == 0, r]         (Mathematica)
r^3 - 3r^2 + 1 = 0                      (Wolfram|Alpha)
roots([1   -3  0  1])                    (Matlab)

(In the Matlab command, one enters the polynomial’s vector of coefficients, listed in descending order.) However we find these approximate roots, it follows that a general solution of the differential equation in (1) is given (approximately) by

(3)y(x)=c1e(0.5321)x+c2e(0.6527)x+c3e(2.8794)x.

Use calculator or computer methods like those indicated here to find general solutions (in approximate numerical form) of the following differential equations.

  1. y(3)3y+y=0

  2. y(3)+3y3y=0

  3. y(3)+y+y=0

  4. y(3)+3y+5y=0

  5. y(4)+2y(3)3y=0

  6. y(4)+3y4y=0

5.4 Mechanical Vibrations

The motion of a mass attached to a spring serves as a relatively simple example of the vibrations that occur in more complex mechanical systems. For many such systems, the analysis of these vibrations is a problem in the solution of linear differential equations with constant coefficients.

We consider a body of mass m attached to one end of an ordinary spring that resists compression as well as stretching; the other end of the spring is attached to a fixed wall, as shown in Fig. 5.4.1. Assume that the body rests on a frictionless horizontal plane, so that it can move only back and forth as the spring compresses and stretches. Denote by x the distance of the body from its equilibrium position—its position when the spring is unstretched. We take x>0 when the spring is stretched, and thus x<0 when it is compressed.

FIGURE 5.4.1.

A mass–spring–dashpot system.

According to Hooke’s law, the restorative force FS that the spring exerts on the mass is proportional to the distance x that the spring has been stretched or compressed. Because this is the same as the displacement x of the mass m from its equilibrium position, it follows that

(1)FS=kx.

The positive constant of proportionality k is called the spring constant. Note that FS and x have opposite signs: FS<0 when x>0, FS>0 when x<0.

Figure 5.4.1 shows the mass attached to a dashpot—a device, like a shock absorber, that provides a force directed opposite to the instantaneous direction of motion of the mass m. We assume the dashpot is so designed that this force FR is proportional to the velocity v=dx/dt of the mass; that is,

(2)FR=cv=cdxdt.

The positive constant c is the damping constant of the dashpot. More generally, we may regard Eq. (2) as specifying frictional forces in our system (including air resistance to the motion of m).

If, in addition to the forces FS and FR, the mass is subjected to a given external force FE=F(t), then the total force acting on the mass is F=FS+FR+FE. Using Newton’s law

F=ma=md2xdt2=mx,

we obtain the second-order linear differential equation

(3)mx+cx+kx=F(t)

that governs the motion of the mass.

If there is no dashpot (and we ignore all frictional forces), then we set c=0 in Eq. (3) and call the motion undamped; it is damped motion if c>0. If there is no external force, we replace F(t) with 0 in Eq. (3). We refer to the motion as free in this case and forced in the case F(t)0. Thus the homogeneous equation

(4)mx+cx+kx=0

describes free motion of a mass on a spring with dashpot but with no external forces applied. We will defer discussion of forced motion until Section 5.6.

For an alternative example, we might attach the mass to the lower end of a spring that is suspended vertically from a fixed support, as in Fig. 5.4.2. In this case the weight W=mg of the mass would stretch the spring a distance s0 determined by Eq. (1) with FS=W and x=s0. That is, mg=ks0, so that s0=mg/k. This gives the static equilibrium position of the mass. If y denotes the displacement of the mass in motion, measured downward from its static equilibrium position, then we ask you to show in Problem 9 that y satisfies Eq. (3); specifically, that

(5)my+cy+ky=F(t)

FIGURE 5.4.2.

A mass suspended vertically from a spring.

if we include damping and external forces (meaning those other than gravity).

The Simple Pendulum

The importance of the differential equation that appears in Eqs. (3) and (5) stems from the fact that it describes the motion of many other simple mechanical systems. For example, a simple pendulum consists of a mass m swinging back and forth on the end of a string (or better, a massless rod) of length L, as shown in Fig. 5.4.3. We may specify the position of the mass at time t by giving the counterclockwise angle θ=θ(t) that the string or rod makes with the vertical at time t. To analyze the motion of the mass m, we will apply the law of the conservation of mechanical energy, according to which the sum of the kinetic energy and the potential energy of m remains constant.

FIGURE 5.4.3.

The simple pendulum.

The distance along the circular arc from 0 to m is s=Lθ, so the velocity of the mass is v=ds/dt=L(dθ/dt), and therefore its kinetic energy is

T=12mv2=12m(dsdt)2=12mL2(dθdt)2.

We next choose as reference point the lowest point O reached by the mass (see Fig. 5.4.3). Then its potential energy V is the product of its weight mg and its vertical height h=L(1cos θ) above O, so

V=mgL(1cosθ).

The fact that the sum of T and V is a constant C therefore gives

12mL2(dθdt)2+mgL(1cosθ)=C.

We differentiate both sides of this identity with respect to t to obtain

mL2(dθdt)(d2θdt2)+mgL(sinθ)dθdt=0,

so

(6)d2θdt2+gLsinθ=0

after removal of the common factor mL2(dθ/dt). This differential equation can be derived in a seemingly more elementary manner using the familiar second law F=ma of Newton (applied to tangential components of the acceleration of the mass and the force acting on it). However, derivations of differential equations based on conservation of energy are often seen in more complex situations where Newton’s law is not so directly applicable, and it may be instructive to see the energy method in a simpler application like the pendulum.

Now recall that if θ is small, then sin θθ (this approximation obtained by retaining just the first term in the Taylor series for sin θ). In fact, sin θ and θ agree to two decimal places when |θ| is at most π/12 (that is, 15°). In a typical pendulum clock, for example, θ would never exceed 15°. It therefore seems reasonable to simplify our mathematical model of the simple pendulum by replacing sin θ with θ in Eq. (6). If we also insert a term cθ to account for the frictional resistance of the surrounding medium, the result is an equation in the form of Eq. (4):

(7)θ+cθ+kθ=0

where k=g/L. Note that this equation is independent of the mass m on the end of the rod. We might, however, expect the effects of the discrepancy between θ and sin θ to accumulate over a period of time, so that Eq. (7) will probably not describe accurately the actual motion of the pendulum over a long period of time.

In the remainder of this section, we first analyze free undamped motion and then free damped motion.

Free Undamped Motion

If we have only a mass on a spring, with neither damping nor external force, then Eq. (3) takes the simpler form

(8)mx+kx=0.

It is convenient to define

(9)ω0=km

and rewrite Eq. (8) as

(8′)x+ω02x=0.

The general solution of Eq. (8′) is

(10)x(t)=Acosω0t+Bsinω0t.

To analyze the motion described by this solution, we choose constants C and α so that

(11)C=A2+B2,cosα=AC,andsinα=BC,

as indicated in Fig. 5.4.4. Note that, although tanα=B/A, the angle α is not given by the principal branch of the inverse tangent function (which gives values only in the interval π/2<x<π/2). Instead, α is the angle between 0 and 2π whose cosine and sine have the signs given in (11), where either A or B or both may be negative. Thus

α={tan1(B/A)if A>0, B>0 (first quadrant),π+tan1(B/A)if A<0 (second or third quadrant),2π+tan1(B/A)if A>0, B<0 (fourth quadrant),

FIGURE 5.4.4.

The angle α.

where tan1(B/A) is the angle in (π/2,π/2) given by a calculator or computer.

In any event, from (10) and (11) we get

x(t)=C(ACcosω0t+BCsinω0t)=C(cosαcosω0t+sinαsinω0t).

With the aid of the cosine addition formula, we find that

(12)x(t)=C cos(ω0tα).

Thus the mass oscillates to and fro about its equilibrium position with

  1. Amplitude C,

  2. Circular frequency ω0, and

  3. Phase angle α.

Such motion is called simple harmonic motion.

If time t is measured in seconds, the circular frequency ω0 has dimensions of radians per second (rad/s). The period of the motion is the time required for the system to complete one full oscillation, so is given by

(13)T=2πω0

seconds; its frequency is

(14)v=1T=ω02π

in hertz (Hz), which measures the number of complete cycles per second. Note that frequency is measured in cycles per second, whereas circular frequency has the dimensions of radians per second.

FIGURE 5.4.5.

Simple harmonic motion.

A typical graph of a simple harmonic position function

x(t)=C cos(ω0tα)=C cos(ω0(tαω0))=C cos(ω0(tδ))

is shown in Fig. 5.4.5, where the geometric significance of the amplitude C, the period T, and the time lag

δ=αω0

are indicated.

If the initial position x(0)=x0 and initial velocity x(0)=v0 of the mass are given, we first determine the values of the coefficients A and B in Eq. (10), then find the amplitude C and phase angle α by carrying out the transformation of x(t) to the form in Eq. (12), as indicated previously.

Example 1

Undamped mass-spring system A body with mass m=12 kilogram (kg) is attached to the end of a spring that is stretched 2 meters (m) by a force of 100 newtons (N). It is set in motion with initial position x0=1 (m) and initial velocity v0=5 (m/s). (Note that these initial conditions indicate that the body is displaced to the right and is moving to the left at time t=0.) Find the position function of the body as well as the amplitude, frequency, period of oscillation, and time lag of its motion.

Solution

The spring constant is k=(100N)/(2m)=50 (N/m), so Eq. (8) yields 12x+50x=0; that is,

x+100x=0.

Consequently, the circular frequency of the resulting simple harmonic motion of the body will be ω0=100=10 (rad/s). Hence it will oscillate with period

T=2πω0=2π100.6283 s

and with frequency

v=1T=ω02π=102π1.5915 Hz.

We now impose the initial conditions x(0)=1 and x(0)=5 on the position function

x(t)=A cos 10t+B sin 10twithx(t)=10A sin 10t+10B cos 10t.

It follows readily that A=1 and B=12, so the position function of the body is

x(t)=cos 10t12sin 10t.

Hence its amplitude of motion is

C=(1)2+(12)2=125 m.

To find the time lag, we write

x(t)=52(25cos 10t15sin 10t)=52cos(10tα),

where the phase angle α satisfies

cosα=25>0andsinα=15<0.

Hence α is the fourth-quadrant angle

α=2π+tan1(1/52/5)=2πtan1(12)5.8195,

and the time lag of the motion is

δ=αω00.5820 s.

With the amplitude and approximate phase angle shown explicitly, the position function of the body takes the form

x(t)125cos(10t5.8195),

and its graph is shown in Fig. 5.4.6.

FIGURE 5.4.6.

Graph of the position function x(t)C cos(ω0tα) in Example 1, with amplitude C1.118, period T0.628, and time δ0.582.

Free Damped Motion

With damping but no external force, the differential equation we have been studying takes the form mx+cx+kx=0; alternatively,

(15)x+2px+ω02x=0,

where ω0=k/m is the corresponding undamped circular frequency and

(16)p=c2m>0.

The characteristic equation r2+2pr+ω02=0 of Eq. (15) has roots

(17)r1, r2=p±(p2ω02)1/2

that depend on the sign of

(18)p2ω02=c24m2km=c24km4m2.

The critical damping ccr is given by ccr=4km, and we distinguish three cases, according as c>ccr, c=ccr, or c<ccr.

Overdamped Case: c>ccr (c2>4k m). Because c is relatively large in this case, we are dealing with a strong resistance in comparison with a relatively weak spring or a small mass. Then (17) gives distinct real roots r1 and r2, both of which are negative. The position function has the form

(19)x(t)=c1er1t+c2er2t.

It is easy to see that x(t)0 as t+ and that the body settles to its equilibrium position without any oscillations (Problem 29). Figure 5.4.7 shows some typical graphs of the position function for the overdamped case; we chose x0 a fixed positive number and illustrated the effects of changing the initial velocity v0. In every case the would-be oscillations are damped out.

FIGURE 5.4.7.

Overdamped motion: x(t)=c1er1t+c2er2t with r1<0 and r2<0. Solution curves are graphed with the same initial position x0 and different initial velocities.

Critically Damped Case: c=ccr (c2=4k m). In this case, (17) gives equal roots r1=r2=p of the characteristic equation, so the general solution is

(20)x(t)=ept(c1+c2t).

Because ept>0 and c1+c2t has at most one positive zero, the body passes through its equilibrium position at most once, and it is clear that x(t)0 as t+. Some graphs of the motion in the critically damped case appear in Fig. 5.4.8, and they resemble those of the overdamped case (Fig. 5.4.7). In the critically damped case, the resistance of the dashpot is just large enough to damp out any oscillations, but even a slight reduction in resistance will bring us to the remaining case, the one that shows the most dramatic behavior.

FIGURE 5.4.8.

Critically damped motion: x(t)=(c1+c2t)ept with p>0. Solution curves are graphed with the same initial position x0 and different initial velocities.

Underdamped Case: c<ccr (c2<4k m). The characteristic equation now has two complex conjugate roots p±iω02p2, and the general solution is

(21)x(t)=ept(A cos ω1t+B sin ω1t),

where

(22)ω1=ω02p2=4k mc22m.

Using the cosine addition formula as in the derivation of Eq. (12), we may rewrite Eq. (20) as

x(t)=Cept(ACcosω1t+BCsinω1t),

so

(23)x(t)=Ceptcos(ω1tα)

where

C=A2+B2,cosα=AC,andsinα=BC.

The solution in (21) represents exponentially damped oscillations of the body around its equilibrium position. The graph of x(t) lies between the “amplitude envelope” curves x=Cept and x=Cept and touches them when ω1tα is an integral multiple of π. The motion is not actually periodic, but it is nevertheless useful to call ω1 its circular frequency (more properly, its pseudofrequency), T1=2π/ω1 its pseudoperiod of oscillation, and Cept its time-varying amplitude. Most of these quantities are shown in the typical graph of underdamped motion in Fig. 5.4.9. Note from Eq. (22) that in this case ω1 is less than the undamped circular frequency ω0, so T1 is larger than the period T of oscillation of the same mass without damping on the same spring. Thus the action of the dashpot has at least two effects:

FIGURE 5.4.9.

Underdamped oscillations: x(t)=Ceptcos(ω1tα).

  1. It exponentially damps the oscillations, in accord with the time-varying amplitude.

  2. It slows the motion; that is, the dashpot decreases the frequency of the motion.

As the following example illustrates, damping typically also delays the motion further—that is, increases the time lag—as compared with undamped motion with the same initial conditions.

Example 2

Damped system The mass and spring of Example 1 are now attached also to a dashpot that provides 1 N of resistance for each meter per second of velocity. The mass is set in motion with the same initial position x(0)=1 and initial velocity x(0)=5 as in Example 1. Now find the position function of the mass, its new frequency and pseudoperiod of motion, its new time lag, and the times of its first four passages through the initial position x=0.

Solution

Rather than memorizing the various formulas given in the preceding discussion, it is better practice in a particular case to set up the differential equation and then solve it directly. Recall that m=12 and k=50; we are now given c=1 in mks units. Hence Eq. (4) is 12x+x+50x=0; that is,

x+2x+100x=0.

The characteristic equation r2+2r+100=(r+1)2+99=0 has roots r1, r2=1±99i, so the general solution is

(24)x(t)=et(A cos 99t+B sin 99t).

Consequently, the new circular (pseudo)frequency is ω1=999.9499 (as compared with ω0=10 in Example 1). The new (pseudo)period and frequency are

T1=2πω1=2π990.6315 s

and

v1=1T1=ω12π=992π1.5836 Hz

(as compared with T0.6283<T1 and ν1.5915>ν1 in Example 1).

We now impose the initial conditions x(0)=1 and x(0)=5 on the position function in (23) and the resulting velocity function

x(t)=et(A cos 99t+B sin 99t)+99et(A sin99t+B cos 99t).

It follows that

x(0)=A=1andx(0)=A+B99=5,

whence we find that A=1 and B=4/99. Thus the new position function of the body is

x(t)=et(cos99t499sin99t).

Hence its time-varying amplitude of motion is

C1et=(1)2+(499)2et=11599et.

We therefore write

x(t)=11599et(99115cos99t4115sin99t)=11599etcos(99tα1),

where the phase angle α1 satisfies

cosα1=99115>0andsinα1=4115<0.

Hence α1 is the fourth-quadrant angle

α1=2π+tan1(4/11599/115)=2πtan1(499)5.9009,

and the time lag of the motion is

δ1=α1ω10.5931 s

(as compared with δ0.5820<δ1 in Example 1). With the time-varying amplitude and approximate phase angle shown explicitly, the position function of the mass takes the form

(25)x(t)11599etcos(99t5.9009),

and its graph is the damped exponential that is shown in Fig. 5.4.10 (in comparison with the undamped oscillations of Example 1).

From (24) we see that the mass passes through its equilibrium position x=0 when cos(ω1tα1)=0, and thus when

ω1tα1=3π2,π2,π2,3π2,;

that is, when

t=δ13π2ω1,δ1π2ω1,δ1+π2ω1,δ1+3π2ω1,.

We see similarly that the undamped mass of Example 1 passes through equilibrium when

t=δ03π2ω0,δ0π2ω0,δ0+π2ω0,δ0+3π2ω0,.

The following table compares the first four values t1,t2,t3,t4 we calculate for the undamped and damped cases, respectively.

n 1 2 3 4
tn (undamped) 0.1107 0.4249 0.7390 1.0532
tn (damped) 0.1195 0.4352 0.7509 1.0667

Accordingly, in Fig. 5.4.11 (where only the first three equilibrium passages are shown) we see the damped oscillations lagging slightly behind the undamped ones.

FIGURE 5.4.10.

Graphs of the position function x(t)=C1etcos(ω1tα1) of Example 2 (damped oscillations), the position function x(t)=C cos(ω0tα) of Example 1 (undamped oscillations), and the envelope curves x(t)=±C1et.

FIGURE 5.4.11.

Graphs on the interval 0t0.8 illustrating the additional delay associated with damping.

5.4 Problems

  1. Determine the period and frequency of the simple harmonic motion of a 4-kg mass on the end of a spring with spring constant 16 N/m.

  2. Determine the period and frequency of the simple harmonic motion of a body of mass 0.75 kg on the end of a spring with spring constant 48 N/m.

  3. A mass of 3 kg is attached to the end of a spring that is stretched 20 cm by a force of 15 N. It is set in motion with initial position x0=0 and initial velocity v0=10 m/s. Find the amplitude, period, and frequency of the resulting motion.

  4. A body with mass 250 g is attached to the end of a spring that is stretched 25 cm by a force of 9 N. At time t=0 the body is pulled 1 m to the right, stretching the spring, and set in motion with an initial velocity of 5 m/s to the left. (a) Find x(t) in the form Ccos(ω0tα). (b) Find the amplitude and period of motion of the body.

Simple Pendulum

In Problems 5 through 8, assume that the differential equation of a simple pendulum of length L is Lθ+gθ=0, where g=GM/R2 is the gravitational acceleration at the location of the pendulum (at distance R from the center of the earth; M denotes the mass of the earth).

  1. Two pendulums are of lengths L1 and L2 and—when located at the respective distances R1 and R2 from the center of the earth—have periods p1 and p2. Show that

    p1p2=R1L1R2L2.
  2. A certain pendulum keeps perfect time in Paris, where the radius of the earth is R=3956 (mi). But this clock loses 2 min 40 s per day at a location on the equator. Use the result of Problem 5 to find the amount of the equatorial bulge of the earth.

  3. A pendulum of length 100.10 in., located at a point at sea level where the radius of the earth is R=3960 (mi), has the same period as does a pendulum of length 100.00 in. atop a nearby mountain. Use the result of Problem 5 to find the height of the mountain.

    FIGURE 5.4.12.

    The buoy of Problem 10.

  4. Most grandfather clocks have pendulums with adjustable lengths. One such clock loses 10 min per day when the length of its pendulum is 30 in. With what length pendulum will this clock keep perfect time?

  5. Derive Eq. (5) describing the motion of a mass attached to the bottom of a vertically suspended spring. (Suggestion: First denote by x(t) the displacement of the mass below the unstretched position of the spring; set up the differential equation for x. Then substitute y=xs0 in this differential equation.)

  6. Floating buoy Consider a floating cylindrical buoy with radius r, height h, and uniform density ρ0.5 (recall that the density of water is 1 g/cm3). The buoy is initially suspended at rest with its bottom at the top surface of the water and is released at time t=0. Thereafter it is acted on by two forces: a downward gravitational force equal to its weight mg=ρπr2hg and (by Archimedes’ principle of buoyancy) an upward force equal to the weight πr2xg of water displaced, where x=x(t) is the depth of the bottom of the buoy beneath the surface at time t (Fig. 5.4.12). Assume that friction is negligible. Conclude that the buoy undergoes simple harmonic motion around its equilibrium position xe=ρh with period p=2πρh/g. Compute p and the amplitude of the motion if ρ=0.5 g/cm3, h=200 cm, and g=980 c m/s2.

  7. Floating buoy A cylindrical buoy weighing 100 lb (thus of mass m=3.125 slugs in ft-lb-s (fps)units) floats in water with its axis vertical (as in Problem 10). When depressed slightly and released, it oscillates up and down four times every 10 s. Find the radius of the buoy.

  8. Hole through the earth Assume that the earth is a solid sphere of uniform density, with mass M and radius R=3960 (mi). For a particle of mass m within the earth at distance r from the center of the earth, the gravitational force attracting m toward the center is Fr=GMrm/r2, where Mr is the mass of the part of the earth within a sphere of radius r (Fig. 5.4.13). (a) Show that Fr=GMmr/R3.

    FIGURE 5.4.13.

    A mass m falling down a hole through the center of the earth (Problem 12).

    (b) Now suppose that a small hole is drilled straight through the center of the earth, thus connecting two antipodal points on its surface. Let a particle of mass m be dropped at time t=0 into this hole with initial speed zero, and let r(t) be its distance from the center of the earth at time t, where we take r<0 when the mass is “below” the center of the earth. Conclude from Newton’s second law and part (a) that r(t)=k2r(t), where k2=GM/R3=g/R. (c) Take g=32.2 ft/s2, and conclude from part (b) that the particle undergoes simple harmonic motion back and forth between the ends of the hole, with a period of about 84 min. (d) Look up (or derive) the period of a satellite that just skims the surface of the earth; compare with the result in part (c). How do you explain the coincidence? Or is it a coincidence? (e) With what speed (in miles per hour) does the particle pass through the center of the earth? (f) Look up (or derive) the orbital velocity of a satellite that just skims the surface of the earth; compare with the result in part (e). How do you explain the coincidence? Or is it a coincidence?

  9. Suppose that the mass in a mass–spring–dashpot system with m=10, c=9, and k=2 is set in motion with x(0)=0 and x(0)=5. (a) Find the position function x(t) and show that its graph looks as indicated in Fig. 5.4.14. (b) Find how far the mass moves to the right before starting back toward the origin.

  10. Suppose that the mass in a mass–spring–dashpot system with m=25, c=10, and k=226 is set in motion with x(0)=20 and x(0)=41. (a) Find the position function x(t) and show that its graph looks as indicated in Fig. 5.4.15. (b) Find the pseudoperiod of the oscillations and the equations of the “envelope curves” that are dashed in the figure.

Free Damped Motion

The remaining problems in this section deal with free damped motion. In Problems 15 through 21, a mass m is attached to both a spring (with given spring constant k) and a dashpot (with given damping constant c). The mass is set in motion with initial position x0 and initial velocity v0. Find the position function x(t) and determine whether the motion is overdamped, critically damped, or underdamped. If it is underdamped, write the position function in the form x(t)=C1ept cos(ω1tα1). Also, find the undamped position function u(t)=C0cos(ω0tα0) that would result if the mass on the spring were set in motion with the same initial position and velocity, but with the dashpot disconnected (so c=0). Finally, construct a figure that illustrates the effect of damping by comparing the graphs of x(t) and u(t).

FIGURE 5.4.14.

The position function x(t) of Problem 13.

  1. m=12, c=3, k=4; x0=2, v0=0

  2. m=3, c=30, k=63; x0=2, v0=2

  3. m=1, c=8, k=16; x0=5, v0=10

  4. m=2, c=12, k=50; x0=0, v0=8

  5. m=4, c=20, k=169; x0=4, v0=16

  6. m=2, c=16, k=40; x0=5, v0=4

  7. m=1, c=10, k=125; x0=6, v0=50

  8. Vertical damped motion A 12-lb weight (mass m=0.375 slugs in fps units) is attached both to a vertically suspended spring that it stretches 6 in. and to a dashpot that provides 3 lb of resistance for every foot per second of velocity. (a) If the weight is pulled down 1 ft below its static equilibrium position and then released from rest at time t=0, find its position function x(t). (b) Find the frequency, time-varying amplitude, and phase angle of the motion.

  9. Car suspension This problem deals with a highly simplified model of a car of weight 3200 lb (mass m=100 slugs in fps units). Assume that the suspension system acts like a single spring and its shock absorbers like a single dashpot, so that its vertical vibrations satisfy Eq. (4) with appropriate values of the coefficients. (a) Find the stiffness coefficient k of the spring if the car undergoes free vibrations at 80 cycles per minute (cycles/min) when its shock absorbers are disconnected. (b) With the shock absorbers connected, the car is set into vibration by driving it over a bump, and the resulting damped vibrations have a frequency of 78 cycles/min. After how long will the time-varying amplitude be 1% of its initial value?

Problems 24 through 34 deal with a mass–spring–dashpot system having position function x(t) satisfying Eq. (4). We write x0=x(0) and v0=x(0) and recall that p=c/(2m), ω02=k/m, and ω12=ω02p2. The system is critically damped, overdamped, or underdamped, as specified in each problem.

FIGURE 5.4.15.

The position function x(t) of Problem 14.

  1. (Critically damped) Show in this case that

    x(t)=(x0+v0t+px0t)ept.
  2. (Critically damped) Deduce from Problem 24 that the mass passes through x=0 at some instant t>0 if and only if x0 and v0+px0 have opposite signs.

  3. (Critically damped) Deduce from Problem 24 that x(t) has a local maximum or minimum at some instant t>0 if and only if v0 and v0+px0 have the same sign.

  4. (Overdamped) Show in this case that

    x(t)=12γ[(v0r2x0)er1t(v0r1x0)er2t],

    where r1,r2=p±p2ω02 and γ=(r1r2)/2>0.

  5. (Overdamped) If x0=0, deduce from Problem 27 that

    x(t)=v0γeptsinh γt.
  6. (Overdamped) Prove that in this case the mass can pass through its equilibrium position x=0 at most once.

  7. (Underdamped) Show that in this case

    x(t)=ept(x0 cos ω1t+v0+px0ω1sin ω1t).
  8. (Underdamped) If the damping constant c is small in comparison with 8mk, apply the binomial series to show that

    ω1ω0(1c28mk).
  9. (Underdamped) Show that the local maxima and minima of

    x(t)=Cept cos(ω1tα)

    occur where

    tan(ω1tα)=pω1.

    Conclude that t2t1=2π/ω1 if two consecutive maxima occur at times t1 and t2.

  10. (Underdamped) Let x1 and x2 be two consecutive local maximum values of x(t). Deduce from the result of Problem 32 that

    lnx1x2=2πpω1.

    The constant Δ=2πp/ω1 is called the logarithmic decrement of the oscillation. Note also that c=mω1Δ/π because p=c/(2m).

Note: The result of Problem 33 provides an accurate method for measuring the viscosity of a fluid, which is an important parameter in fluid dynamics but is not easy to measure directly. According to Stokes’s drag law, a spherical body of radius a moving at a (relatively slow)speed through a fluid of viscosity μ experiences a resistive force FR=6πμav. Thus if a spherical mass on a spring is immersed in the fluid and set in motion, this drag resistance damps its oscillations with damping constant c=6πaμ. The frequency ω1 and logarithmic decrement Δ of the oscillations can be measured by direct observation. The final formula in Problem 33 then gives c and hence the viscosity of the fluid.

  1. (Underdamped) A body weighing 100 lb (mass m=3.125 slugs in fps units) is oscillating attached to a spring and a dashpot. Its first two maximum displacements of 6.73 in. and 1.46 in. are observed to occur at times 0.34 s and 1.17 s, respectively. Compute the damping constant (in pound-seconds per foot) and spring constant (in pounds per foot).

Differential Equations and Determinism

Given a mass m, a dashpot constant c, and a spring constant k, Theorem 2 of Section 5.1 implies that the equation

(26)mx+cx+kx=0

has a unique solution for t0 satisfying given initial conditions x(0)=x0, x(0)=v0. Thus the future motion of an ideal mass–spring–dashpot system is completely determined by the differential equation and the initial conditions. Of course in a real physical system it is impossible to measure the parameters m, c, and k precisely. Problems 35 through 38 explore the resulting uncertainty in predicting the future behavior of a physical system.

  1. Suppose that m=1, c=2, and k=1 in Eq. (26). Show that the solution with x(0)=0 and x(0)=1 is

    x1(t)=tet.
  2. Suppose that m=1 and c=2 but k=1102n. Show that the solution of Eq. (26) with x(0)=0 and x(0)=1 is

    x2(t)=10net sinh 10nt.
  3. Suppose that m=1 and c=2 but that k=1+102n. Show that the solution of Eq. (26) with x(0)=0 and x(0)=1 is

    x3(t)=10net sin 10nt.
  4. Whereas the graphs of x1(t) and x2(t) resemble those shown in Figs. 5.4.7 and 5.4.8, the graph of x3(t) exhibits damped oscillations like those illustrated in Fig. 5.4.9, but with a very long pseudoperiod. Nevertheless, show that for each fixed t>0 it is true that

    limnx2(t)=limnx3(t)=x1(t).

    Conclude that on a given finite time interval the three solutions are in “practical” agreement if n is sufficiently large.

5.5 Nonhomogeneous Equations and Undetermined Coefficients

We learned in Section 5.3 how to solve homogeneous linear equations with constant coefficients, but we saw in Section 5.4 that an external force in a simple mechanical system contributes a nonhomogeneous term to its differential equation. The general nonhomogeneous nth-order linear equation with constant coefficients has the form

(1)any(n)+an1y(n1)++a1y+a0y=f(x).

By Theorem 5 of Section 5.2, a general solution of Eq. (1) has the form

(2)y=yc+yp

where the complementary function yc(x) is a general solution of the associated homogeneous equation

(3)any(n)+an1y(n1)++a1y+a0y=0,

and yp(x) is a particular solution of Eq. (1). Thus our remaining task is to find yp.

The method of undetermined coefficients is a straightforward way of doing this when the given function f(x) in Eq. (1) is sufficiently simple that we can make an intelligent guess as to the general form of yp. For example, suppose that f(x) is a polynomial of degree m. Then, because the derivatives of a polynomial are themselves polynomials of lower degree, it is reasonable to suspect a particular solution

yp(x)=Amxm+Am1xm1++A1x+A0

that is also a polynomial of degree m, but with as yet undetermined coefficients. We may, therefore, substitute this expression for yp into Eq. (1), and then—by equating coefficients of like powers of x on the two sides of the resulting equation—attempt to determine the coefficients A0, A1, , Am so that yp will, indeed, be a particular solution of Eq. (1).

Similarly, suppose that

f(x)=a cos kx+b sin kx.

Then it is reasonable to expect a particular solution of the same form:

yp(x)=A cos kx+B sin kx,

a linear combination with undetermined coefficients A and B. The reason is that any derivative of such a linear combination of cos kx and sin kx has the same form. We may therefore substitute this form of yp in Eq. (1), and then—by equating coefficients of cos kx and sin kx on both sides of the resulting equation—attempt to determine the coefficients A and B so that yp will, indeed, be a particular solution.

It turns out that this approach does succeed whenever all the derivatives of f(x) have the same form as f(x) itself. Before describing the method in full generality, we illustrate it with several preliminary examples.

Example 1

Find a particular solution of y+3y+4y=3x+2.

Solution

Here f(x)=3x+2 is a polynomial of degree 1, so our guess is that

yp(x)=Ax+B.

Then yp=A and yp=0, so yp will satisfy the differential equation provided that

(0)+3(A)+4(Ax+B)=3x+2,

that is,

(4Ax)+(3A+4B)=3x+2

for all x. This will be true if the x-terms and constant terms on the two sides of this equation agree. It therefore suffices for A and B to satisfy the two linear equations 4A=3 and 3A+4B=2 that we readily solve for A=34 and B=116. Thus we have found the particular solution

yp(x)=34x116.

Example 2

Find a particular solution of y4y=2e3x.

Solution

Any derivative of e3x is a constant multiple of e3x, so it is reasonable to try

yp(x)=Ae3x.

Then yp=9Ae3x, so the given differential equation will be satisfied provided that

9Ae3x4(Ae3x)=2e3x;

that is, 5A=2, so that A=25. Thus our particular solution is yp(x)=25e3x.

Example 3

Find a particular solution of 3y+y2y=2cos x.

Solution

A first guess might be yp(x)=Acos x, but the presence of y on the left-hand side signals that we probably need a term involving sin x as well. So we try

yp(x)=A cos x+B sin x,yp(x)=A sin x+B cos x,yp(x)=A cos xB sin x.

Then substitution of yp and its derivatives into the given differential equation yields

3(A cos xB sin x)+(A sin x+B cos x)2(A cos x+B sin x)=2 cos x,

that is (collecting coefficients on the left),

(5A+B)cos x+(A5B)sin x=2 cos x.

This will be true for all x provided that the cosine and sine terms on the two sides of this equation agree. It therefore suffices for A and B to satisfy the two linear equations

5A+B=2,A5B=0

with readily found solution A=513, B=113. Hence a particular solution is

yp(x)=513cos x+113sin x.

The following example, which superficially resembles Example 2, indicates that the method of undetermined coefficients is not always quite so simple as we have made it appear.

Example 4

Find a particular solution of y4y=2e2x.

Solution

If we try yp(x)=Ae2x, we find that

yp4yp=4Ae2x4Ae2x=02e2x.

Thus, no matter how A is chosen, Ae2x cannot satisfy the given nonhomogeneous equation. In fact, the preceding computation shows that Ae2x satisfies instead the associated homogeneous equation. Therefore, we should begin with a trial function yp(x) whose derivative involves both e2x and something else that can cancel upon substitution into the differential equation to leave the e2x term that we need. A reasonable guess is

yp(x)=Axe2x,

for which

yp(x)=Ae2x+2Axe2xandyp(x)=4Ae2x+4Axe2x.

Substitution into the original differential equation yields

(4Ae2x+4Axe2x)4(Axe2x)=2e2x.

The terms involving xe2x obligingly cancel, leaving only 4Ae2x=2e2x, so that A=12. Consequently, a particular solution is

yp(x)=12xe2x.

The General Approach

Our initial difficulty in Example 4 resulted from the fact that f(x)=2e2x satisfies the associated homogeneous equation. Rule 1, given shortly, tells what to do when we do not have this difficulty, and Rule 2 tells what to do when we do have it.

The method of undetermined coefficients applies whenever the function f(x) in Eq. (1) is a linear combination of (finite) products of functions of the following three types:

  1. A polynomial in x;

  2. An exponential function erx; (4)

  3. cos kx or sin kx.

Any such function, for example,

f(x)=(34x2)e5x4x3cos 10x,

has the crucial property that only finitely many linearly independent functions appear as terms (summands) in f(x) and its derivatives of all orders. In Rules 1 and 2 we assume that Ly=f(x) is a nonhomogeneous linear equation with constant coefficients and that f(x) is a function of this kind.

Note that this rule is not a theorem requiring proof; it is merely a procedure to be followed in searching for a particular solution yp. If we succeed in finding yp, then nothing more need be said. (It can be proved, however, that this procedure will always succeed under the conditions specified here.)

In practice we check the supposition made in Rule 1 by first using the characteristic equation to find the complementary function yc, and then write a list of all the terms appearing in f(x) and its successive derivatives. If none of the terms in this list duplicates a term in yc, then we proceed with Rule 1.

Example 5

Find a particular solution of

(5)y+4y=3x3.

Solution

The (familiar) complementary solution of Eq. (5) is

yc(x)=c1 cos 2x+c2 sin 2x.

The function f(x)=3x3 and its derivatives are constant multiples of the linearly independent functions x3, x2, x, and 1. Because none of these appears in yc, we try

yp=Ax3+Bx2+Cx+D,yp=3Ax2+2Bx+C,yp=6Ax+2B.

Substitution in Eq. (5) gives

yp+4yp=(6Ac+2B)+4(Ax3+Bx2+Cx+D)=4Ax3+4Bx2+(6A+4C)x+(2B+4D)=3x3.

We equate coefficients of like powers of x in the last equation to get

4A=3,4B=0,6A+4C=0,2B+4D=0

with solution A=34, B=0, C=98, and D=0. Hence a particular solution of Eq. (5) is

yp(x)=34x398x.

Example 6

Solve the initial value problem

(6)y3y+2y=3ex10 cos 3x;y(0)=1,y(0)=2.

Solution

The characteristic equation r23r+2=0 has roots r=1 and r=2, so the complementary function is

yc(x)=c1ex+c2e2x.

The terms involved in f(x)=3ex10 cos 3x and its derivatives are ex, cos 3x, and sin 3x. Because none of these appears in yc, we try

yp=Aex+B cos 3x+C sin 3x,yp=Aex3B sin 3x+3C cos 3x,yp=Aex9B cos 3x9C sin 3x.

After we substitute these expressions into the differential equation in (6) and collect coefficients, we get

yp3yp+2yp=6Aex+(7B9C)cos 3x+(9B7C) sin 3x=3ex10 cos 3x.

We equate the coefficients of the terms involving ex, those involving cos 3x, and those involving sin 3x. The result is the system

6A=3,7B9C=10,9B7C=0

with solution A=12, B=713, and C=913. This gives the particular solution

yp(x)=12ex+713cos 3x+913sin 3x,

which, however, does not have the required initial values in (6).

To satisfy those initial conditions, we begin with the general solution

y(x)=yc(x)+yp(x)=c1ex+c2e2x+12ex+713cos 3x+913sin 3x,

with derivative

y(x)=c1ex+2c2e2x12ex2113sin 3x+2713cos 3x.

The initial conditions in (6) lead to the equations

y(0)=c1+c2+12+713=1,y(0)=c1+2c212+2713=2

with solution c1=12, c2=613. The desired particular solution is therefore

y(x)=12ex+613e2x+12ex+713cos 3x+913sin 3x.

Example 7

Find the general form of a particular solution of

(7)y(3)+9y=xsinx+x2e2x.

Solution

The characteristic equation r3+9r=0 has roots r=0, r=3i, and r=3i. So the complementary function is

yc(x)=c1+c2cos 3x+c3sin 3x.

The derivatives of the right-hand side in Eq. (7) involve the terms

cos x,sin x,x cos x,x sin x,e2x,xe2x,andx2e2x.

Because there is no duplication with the terms of the complementary function, the trial solution takes the form

yp(x)=A cos x+B sin x+Cx cos x+Dx sin x+Ee2x+Fxe2x+Gx2e2x.

Upon substituting yp in Eq. (7) and equating coefficients of like terms, we get seven equations determining the seven coefficients A, B, C, D, E, F, and G.

The Case of Duplication

Now we turn our attention to the situation in which Rule 1 does not apply: Some of the terms involved in f(x) and its derivatives satisfy the associated homogeneous equation. For instance, suppose that we want to find a particular solution of the differential equation

(8)(Dr)3y=(2x3)erx.

Proceeding as in Rule 1, our first guess would be

(9)yp(x)=Aerx+Bxerx.

This form of yp(x) will not be adequate because the complementary function of Eq. (8) is

(10)yc(x)=c1erx+c2xerx+c3x2erx,

so substitution of (9) in the left-hand side of (8) would yield zero rather than (2x3)erx.

To see how to amend our first guess, we observe that

(Dr)2[(2x3)erx]=[D2(2x3)]erx=0

by Eq. (13) of Section 5.3. If y(x) is any solution of Eq. (8) and we apply the operator (Dr)2 to both sides, we see that y(x) is also a solution of the equation (Dr)5y=0. The general solution of this homogeneous equation can be written as

y(x)=c1erx+c2xerx+c3x2erxyc+Ax3erx+Bx4erxyp.

Thus every solution of our original equation in (8) is the sum of a complementary function and a particular solution of the form

(11)yp(x)=Ax3erx+Bx4erx.

Note that the right-hand side in Eq. (11) can be obtained by multiplying each term of our first guess in (9) by the least positive integral power of x (in this case, x3) that suffices to eliminate duplication between the terms of the resulting trial solution yp(x) and the complementary function yc(x) given in (10). This procedure succeeds in the general case.

To simplify the general statement of Rule 2, we observe that to find a particular solution of the nonhomogeneous linear differential equation

(12)Ly=f1(x)+f2(x),

it suffices to find separately particular solutions Y1(x) and Y2(x) of the two equations

(13)Ly=f1(x)andLy=f2(x),

respectively. For linearity then gives

L[Y1+Y2]=LY1+LY2=f1(x)+f2(x),

and therefore yp=Y1+Y2 is a particular solution of Eq. (12). (This is a type of “superposition principle” for nonhomogeneous linear equations.)

Now our problem is to find a particular solution of the equation Ly=f(x), where f(x) is a linear combination of products of the elementary functions listed in (4). Thus f(x) can be written as a sum of terms each of the form

(14)Pm(x)erx cos kxorPm(x)erx sin kx,

where Pm(x) is a polynomial in x of degree m. Note that any derivative of such a term is of the same form but with both sines and cosines appearing. The procedure by which we arrived earlier at the particular solution in (11) of Eq. (8) can be generalized to show that the following procedure is always successful.

In practice we seldom need to deal with a function f(x) exhibiting the full generality in (14). The table in Fig. 5.5.1 lists the form of yp in various common cases, corresponding to the possibilities m=0, r=0, and k=0.

On the other hand, it is common to have

f(x)=f1(x)+f2(x),

where f1(x) and f2(x) are different functions of the sort listed in the table in Fig. 5.5.1. In this event we take as yp the sum of the trial solutions for f1(x) and f2(x), choosing s separately for each part to eliminate duplication with the complementary function. This procedure is illustrated in Examples 8 through 10.

Example 8

Find a particular solution of

(16)y(3)+y=3ex+4x2.

FIGURE 5.5.1.

Substitutions in the method of undetermined coefficients.

f(x) yp
Pm(x)=b0+b1x+b2x2++bmxma cos k x+b sin k xerx(a cos k x+b sin k x)Pm(x)erxPm(x)(a cos k x+b sin k x) xs(A0+A1x+A2x2++Amxm)xs(A cos k x+B sin k x)xserx(A cos k x+B sin k x)xs(A0+A1x+A2x2++Amxm)erxxs[(A0+A1x++Amxm)coskx+(B0+B1x++Bmxm)sinkx]

Solution

The characteristic equation r3+r2=0 has roots r1=r2=0 and r3=1, so the complementary function is

yc(x)=c1+c2x+c3ex.

As a first step toward our particular solution, we form the sum

(Aex)+(B+Cx+Dx2).

The part Aex corresponding to 3ex does not duplicate any part of the complementary function, but the part B+Cx+Dx2 must be multiplied by x2 to eliminate duplication. Hence we take

yp=Aex+Bx2+Cx3+Dx4,yp=Aex+2Bx+3Cx2+4Dx3,yp=Aex+2B+6Cx+12Dx2,andyp(3)=Aex+6C+24Dx.

Substitution of these derivatives in Eq. (16) yields

2Aex+(2B+6C)+(6C+24D)x+12Dx2=3ex+4x2.

The system of equations

2A=3,2B+6C=0,6C+24D=0,12D=4

has the solution A=32, B=4, C=43, and D=13. Hence the desired particular solution is

yp(x)=32ex+4x243x3+13x4.

Example 9

Determine the appropriate form for a particular solution of

y+6y+13y=e3x cos 2x.

Solution

The characteristic equation r2+6r+13=0 has roots 3±2i, so the complementary function is

yc(x)=e3x(c1 cos 2x+c2 sin 2x).

This is the same form as a first attempt e3x(Acos 2x+Bsin 2x) at a particular solution, so we must multiply by x to eliminate duplication. Hence we would take

yp(x)=e3x(Ax cos 2x+Bx sin 2x).

Example 10

Determine the appropriate form for a particular solution of the fifth-order equation

(D2)3(D2+9)y=x2e2x+x sin 3x.

Solution

The characteristic equation (r2)3(r2+9)=0 has roots r=2, 2, 2, 3i, and 3i, so the complementary function is

yc(x)=c1e2x+c2xe2x+c3x2e2x+c4 cos 3x+c5 sin 3x.

As a first step toward the form of a particular solution, we examine the sum

[(A+Bx+Cx2)e2x]+[(D+Ex)cos 3x+(F+Gx) sin 3x].

To eliminate duplication with terms of yc(x), the first part—corresponding to x2e2x—must be multiplied by x3, and the second part—corresponding to xsin 3x—must be multiplied by x. Hence we would take

yp(x)=(Ax3+Bx4+Cx5)e2x+(Dx+Ex2)cos 3x+(Fx+Gx2)sin 3x.

Variation of Parameters

Finally, let us point out the kind of situation in which the method of undetermined coefficients cannot be used. Consider, for example, the equation

(17)y+y=tan x,

which at first glance may appear similar to those considered in the preceding examples. Not so; the function f(x)=tanx has infinitely many linearly independent derivatives

sec2 x,2 sec2 x tan x,4 sec2 x tan2 x+2 sec4 x,.

Therefore, we do not have available a finite linear combination to use as a trial solution.

We discuss here the method of variation of parameters, which—in principle (that is, if the integrals that appear can be evaluated)—can always be used to find a particular solution of the nonhomogeneous linear differential equation

(18)y(n)+pn1(x)y(n1)++p1(x)y+p0(x)y=f(x),

provided that we already know the general solution

(19)yc=c1y1+c2y2++cnyn

of the associated homogeneous equation

(20)y(n)+pn1(x)y(n1)++p1(x)y+p0(x)y=0.

Here, in brief, is the basic idea of the method of variation of parameters. Suppose that we replace the constants, or parameters, c1, c2, , cn in the complementary function in Eq. (19) with variables: functions u1, u2, , un of x. We ask whether it is possible to choose these functions in such a way that the combination

(21)yp(x)=u1(x)y1(x)+u2(x)y2(x)++un(x)yn(x)

is a particular solution of the nonhomogeneous equation in (18). It turns out that this is always possible.

The method is essentially the same for all orders n2, but we will describe it in detail only for the case n=2. So we begin with the second-order nonhomogeneous equation

(22)L[y]=y+P(x)y+Q(x)y=f(x)

with complementary function

(23)yc(x)=c1y1(x)+c2y2(x)

on some open interval I where the functions P and Q are continuous. We want to find functions u1 and u2 such that

(24)yp(x)=u1(x)y1(x)+u2(x)y2(x)

is a particular solution of Eq. (22).

One condition on the two functions u1 and u2 is that L[yp]=f(x). Because two conditions are required to determine two functions, we are free to impose an additional condition of our choice. We will do so in a way that simplifies the computations as much as possible. But first, to impose the condition L[yp]=f(x), we must compute the derivatives yp and yp. The product rule gives

yp=(u1y1+u2y2)+(u1y1+u2y2).

To avoid the appearance of the second derivatives u1 and u2, the additional condition that we now impose is that the second sum here must vanish:

(25)u1y1+u2y2=0.

Then

(26)yp=u1y1+u2y2,

and the product rule gives

(27)yp=(u1y1+u2y2)+(u1y1+u2y2).

But both y1 and y2 satisfy the homogeneous equation

y+Py+Qy=0

associated with the nonhomogeneous equation in (22), so

(28)yi=PyiQyi

for i=1, 2. It therefore follows from Eq. (27) that

yp=(u1y1+u2y2)P(u1y1+u2y2)Q(u1y1+u2y2).

In view of Eqs. (24) and (26), this means that

yp=(u1y1+u2y2)PypQyp;

hence

(29)L[yp]=u1y1+u2y2.

The requirement that yp satisfy the nonhomogeneous equation in (22)—that is, that L[yp]=f(x)—therefore implies that

(30)u1y1+u2y2=f(x).

Finally, Eqs. (25) and (30) determine the functions u1 and u2 that we need. Collecting these equations, we obtain a system

(31)u1y1+u2y2=0,u1y1+u2y2=f(x)

of two linear equations in the two derivatives u1 and u2. Note that the determinant of coefficients in (31) is simply the Wronskian W(y1,y2). Once we have solved the equations in (31) for the derivatives u1 and u2, we integrate each to obtain the functions u1 and u2 such that

(32)yp=u1y1+u2y2

is the desired particular solution of Eq. (22). In Problem 63 we ask you to carry out this process explicitly and thereby verify the formula for yp(x) in the following theorem.

Example 11

Find a particular solution of the equation y+y=tanx.

Solution

The complementary function is yc(x)=c1cos x+c2sin x, and we could simply substitute directly in Eq. (33). But it is more instructive to set up the equations in (31) and solve for u1 and u2, so we begin with

y1=cos x,y2=sin x,y1=sin x,y2=cos x.

Hence the equations in (31) are

(u1)(cos x)+(u2)(sin x)=0,(u1)(sin x)+(u2)(cos x)=tan x.

We easily solve these equations for

u1=sin x tan x=sin2 xcos x=cos xsecx,u2=cos x tan x=sin x.

Hence we take

u1=(cosxsecx)dx=sinxln|sec x+tan x|

and

u2=sin x dx=cos x.

(Do you see why we choose the constants of integration to be zero?) Thus our particular solution is

yp(x)=u1(x)y1(x)+u2(x)y2(x)=(sin xln|sec x+tanx|)cos x+(cos x)(sin x);

that is,

yp(x)=(cos x)ln|sec x+tan x|.

5.5 Problems

In Problems 1 through 20, find a particular solution yp of the given equation. In all these problems, primes denote derivatives with respect to x.

  1. y+16y=e3x

  2. yy2y=3x+4

  3. yy6y=2 sin 3x

  4. 4y+4y+y=3xex

  5. y+y+y=sin2 x

  6. 2y+4y+7y=x2

  7. y4y=sinh x

  8. y4y=cosh 2x

  9. y+2y3y=1+xex

  10. y+9y=2 cos 3x+3 sin 3x

  11. y(3)+4y=3x1

  12. y(3)+y=2sin x

  13. y+2y+5y=ex sin x

  14. y(4)2y+y=xex

  15. y(5)+5y(4)y=17

  16. y+9y=2x2e3x+5

  17. y+y=sin x+xcos x

  18. y(4)5y+4y=exxe2x

  19. y(5)+2y(3)+2y=3x21

  20. y(3)y=ex+7

In Problems 21 through 30, set up the appropriate form of a particular solution yp, but do not determine the values of the coefficients.

  1. y2y+2y=ex sin x

  2. y(5)y(3)=ex+2x25

  3. y+4y=3xcos 2x

  4. y(3)y12y=x2xe3x

  5. y+3y+2y=x(exe2x)

  6. y6y+13y=xe3x sin 2x

  7. y(4)+5y+4y=sin x+cos 2x

  8. y(4)+9y=(x2+1)sin 3x

  9. (D1)3(D24)y=xex+e2x+e2x

  10. y(4)2y+y=x2 cos x

Solve the initial value problems in Problems 31 through 40.

  1. y+4y=2x; y(0)=1, y(0)=2

  2. y+3y+2y=ex; y(0)=0, y(0)=3

  3. y+9y=sin 2x; y(0)=1, y(0)=0

  4. y+y=cos x; y(0)=1, y(0)=1

  5. y2y+2y=x+1; y(0)=3, y(0)=0

  6. y(4)4y=x2; y(0)=y(0)=1, y(0)=y(3)(0)=1

  7. y(3)2y+y=1+xex; y(0)=y(0)=0, y(0)=1

  8. y+2y+2y=sin 3x; y(0)=2, y(0)=0

  9. y(3)+y=x+ex; y(0)=1, y(0)=0, y(0)=1

  10. y(4)y=5; y(0)=y(0)=y(0)=y(3)(0)=0

  11. Find a particular solution of the equation

    y(4)y(3)yy2y=8x5.
  12. Find the solution of the initial value problem consisting of the differential equation of Problem 41 and the initial conditions

    y(0)=y(0)=y(0)=y(3)(0)=0.
    1. Write

      cos 3x+i sin 3x=e3ix=(cos x+i sin x)3

      by Euler’s formula, expand, and equate real and imaginary parts to derive the identities

      cos3 x=34cos x+14cos 3x,sin3 x=34sin x14sin 3x.
    2. Use the result of part (a) to find a general solution of

      y+4y=cos3 x.

Use trigonometric identities to find general solutions of the equations in Problems 44 through 46.

  1. y+y+y=sin xsin 3x

  2. y+9y=sin4 x

  3. y+y=xcos3 x

In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation.

  1. y+3y+2y=4ex

  2. y2y8y=3e2x

  3. y4y+4y=2e2x

  4. y4y=sinh2x

  5. y+4y=cos 3x

  6. y+9y=sin 3x

  7. y+9y=2sec3x

  8. y+y=csc2 x

  9. y+4y=sin2 x

  10. y4y=xex

  11. You can verify by substitution that yc=c1x+c2x1 is a complementary function for the nonhomogeneous second-order equation

    x2y+xyy=72x5.

    But before applying the method of variation of parameters, you must first divide this equation by its leading coefficient x2 to rewrite it in the standard form

    y+1xy1x2y=72x3.

    Thus f(x)=72x3 in Eq. (22). Now proceed to solve the equations in (31) and thereby derive the particular solution yp=3x5.

In Problems 58 through 62, a nonhomogeneous second-order linear equation and a complementary function yc are given. Apply the method of Problem 57 to find a particular solution of the equation.

  1. x2y4xy+6y=x3; yc=c1x2+c2x3

  2. x2y3xy+4y=x4; yc=x2(c1+c2ln x)

  3. 4x2y4xy+3y=8x4/3; yc=c1x+c2x3/4

  4. x2y+xy+y=ln x; yc=c1cos(ln x)+c2sin(ln x)

  5. (x21)y2xy+2y=x21; yc=c1x+c2(1+x2)

  6. Carry out the solution process indicated in the text to derive the variation of parameters formula in (33) from Eqs. (31) and (32).

  7. Apply the variation of parameters formula in (33) to find the particular solution yp(x)=xcos x of the nonhomogeneous equation y+y=2sin x.

5.5 Application Automated Variation of Parameters

The variation of parameters formula in (33) is especially amenable to implementation in a computer algebra system when the indicated integrals would be too tedious or inconvenient for manual evaluation. For example, suppose that we want to find a particular solution of the nonhomogeneous equation

y+y=tan x

of Example 11, with complementary function yc(x)=c1cos x+c2sin x. Then the Maple commands

y1 := cos(x):
y2 := sin(x):
f := tan(x):
W := y1*diff(y2,x) - y2*diff(y1,x):
W := simplify(W):
yp := -y1*int(y2*f/W,x) + y2*int(y1*f/W,x):
simplify(yp);

implement (33) and produce the result

yp(x)=(cos x)ln(1+sin xcos x)

equivalent to the result yp(x)=(cos x)ln (secx+tanx) found in Example 11. The analogous Mathematica commands

y1 = Cos[x];
y2 = Sin[x];
f = Tan[x];
W = y1*D[y2,x] - y2*D[y1,x]   //  Simplify
yp = -y1*Integrate[y2*f/W,x] + y2*Integrate[y1*f/W,x];
Simplify[yp]

produce the result

yp(x)=(cos x)ln(cos(x/2)+sin(x/2)cos(x/2)sin(x/2)),

which (by the usual difference-of-squares technique) also is equivalent to the result found in Example 11.

To solve similarly a second-order linear equation y+P(x)y+Q(x)y=f(x) whose complementary function yc(x)=c1y1(x)+c2y2(x) is known, we need only insert the corresponding definitions of y1(x), y2(x), and f(x) in the initial lines shown here. Find in this way the indicated particular solution yp(x) of the nonhomogeneous equations in Problems 1 through 6.

1. y+y=2 sin x yp(x)=xcos x
2. y+y=4x sin x yp(x)=xsin xx2 cos x
3. y+y=12x2 sin x yp(x)=3x2 sin x+(3x2x3) cos x
4. y2y+2y=2ex sin x yp(x)=xex cos x
5. y2y+2y=4xex sin x yp(x)=ex(xsin xx2 cos x)
6. y2y+2y=12x2ex sin x yp(x)=ex[3x2 sin x+(3x2x3) cos x]

5.6 Forced Oscillations and Resonance

In Section 5.4 we derived the differential equation

(1)mx+cx+kx=F(t)

that governs the one-dimensional motion of a mass m that is attached to a spring (with constant k) and a dashpot (with constant c) and is also acted on by an external force F(t). Machines with rotating components commonly involve mass-spring systems (or their equivalents) in which the external force is simple harmonic:

(2)F(t)=F0 cos ωtorF(t)=F0 sin ωt,

where the constant F0 is the amplitude of the periodic force and ω is its circular frequency.

For an example of how a rotating machine component can provide a simple harmonic force, consider the cart with a rotating vertical flywheel shown in Fig. 5.6.1. The cart has mass mm0, not including the flywheel of mass m0. The centroid of the flywheel is off center at a distance a from its center, and its angular speed is ω radians per second. The cart is attached to a spring (with constant k) as shown. Assume that the centroid of the cart itself is directly beneath the center of the flywheel, and denote by x(t) its displacement from its equilibrium position (where the spring is unstretched). Figure 5.6.1 helps us to see that the displacement x̲ of the centroid of the combined cart plus flywheel is given by

x¯=(mm0)x+m0(x+a cos ωt)m=x+m0amcos ωt.

FIGURE 5.6.1.

The cart-with-flywheel system.

Let us ignore friction and apply Newton’s second law mx̲=kx, because the force exerted by the spring is kx. We substitute for x̲ in the last equation to obtain

mxm0aω2 cos ωt=kx;

that is,

(3)mx+k x=m0aω2 cos ωt.

Thus the cart with its rotating flywheel acts like a mass on a spring under the influence of a simple harmonic external force with amplitude F0=m0aω2. Such a system is a reasonable model of a front-loading washing machine with the clothes being washed loaded off center. This illustrates the practical importance of analyzing solutions of Eq. (1) with external forces as in (2).

Undamped Forced Oscillations

To study undamped oscillations under the influence of the external force F(t)=F0cos ωt, we set c=0 in Eq. (1), and thereby begin with the equation

(4)mx+k x=F0 cos ωt

whose complementary function is xc=c1cosω0t+c2sinω0t. Here

ω0=km

(as in Eq. (9) of Section 5.4) is the (circular) natural frequency of the mass–spring system. The fact that the angle ω0t is measured in (dimensionless) radians reminds us that if t is measured in seconds (s), then ω0 is measured in radians per second—that is, in inverse seconds (s1). Also recall from Eq. (14) in Section 5.4 that division of a circular frequency ω by the number 2π of radians in a cycle gives the corresponding (ordinary) frequency ν=ω/2π in Hz (hertz = cycles per second).

Let us assume initially that the external and natural frequencies are unequal: ωω0. We substitute xp=Acos ωt in Eq. (4) to find a particular solution. (No sine term is needed in xp because there is no term involving x on the left-hand side in Eq. (4).) This gives

mω2 A cos ωt+k A cos ωt=F0 cos ωt,

so

(5)A=F0kmω2=F0/mω02ω2,

and thus

(6)xp(t)=F0/mω02ω2cos ωt.

Therefore, the general solution x=xc+xp is given by

(7)x(t)=c1 cos ω0t+c2 sin ω0t+F0/mω02ω2cos ωt,

where the constants c1 and c2 are determined by the initial values x(0) and x(0). Equivalently, as in Eq. (12) of Section 5.4, we can rewrite Eq. (7) as

(8)x(t)=C cos(ω0tα)+F0/mω02ω2cos ωt,

so we see that the resulting motion is a superposition of two oscillations, one with natural circular frequency ω0, the other with the frequency ω of the external force.

Example 1

Suppose that m=1, k=9, F0=80, and ω=5, so the differential equation in (4) is

x+9x=80 cos 5t.

Find x(t) if x(0)=x(0)=0.

Solution

Here the natural frequency ω0=3 and the frequency ω=5 of the external force are unequal, as in the preceding discussion. First we substitute xp=Acos 5t in the differential equation and find that 25A+9A=80, so that A=5. Thus a particular solution is

xp(t)=5 cos 5t.

The complementary function is xc=c1cos 3t+c2sin 3t, so the general solution of the given nonhomogeneous equation is

x(t)=c1cos 3t+c2sin 3t5 cos 5t,

with derivative

x(t)=3c1sin 3t+3c2cos 3t+25 sin 5t.

The initial conditions x(0)=0 and x(0)=0 now yield c1=5 and c2=0, so the desired particular solution is

x(t)=5 cos 3t5 cos 5t.

As indicated in Fig. 5.6.2, the period of x(t) is the least common multiple 2π of the periods 2π/3 and 2π/5 of the two cosine terms.

FIGURE 5.6.2.

The response x(t)=5 cos 3t5 cos 5t in Example 1.

Beats

If we impose the initial conditions x(0)=x(0)=0 on the solution in (7), we find that

c1=F0m(ω02ω2)andc2=0,

so the particular solution is

(9)x(t)=F0m(ω02ω2)(cosωtcosω0t).

The trigonometric identity 2sin Asin B=cos(AB)cos(A+B), applied with A=12(ω0+ω)t and B=12(ω0ω)t, enables us to rewrite Eq. (9) in the form

(10)x(t)=F0m(ω02ω2)sin 12(ω0ω)t sin12(ω0+ω)t.

Suppose now that ωω0, so that ω0+ω is very large in comparison with |ω0ω|. Then sin12(ω0+ω)t is a rapidly varying function, whereas sin12(ω0ω)t is a slowly varying function. We may therefore interpret Eq. (10) as a rapid oscillation with circular frequency 12(ω0+ω),

x(t)=A(t) sin 12(ω0+ω)t,

but with a slowly varying amplitude

A(t)=2F0m(ω02ω2)sin 12(ω0ω)t.

Example 2

With m=0.1, F0=50, ω0=55, and ω=45, Eq. (10) gives

x(t)=sin 5t sin 50t.

Figure 5.6.3 shows the corresponding oscillation of frequency 12(ω0+ω)=50 that is “modulated” by the amplitude function A(t)=sin 5t of frequency 12(ω0ω)=5.

FIGURE 5.6.3.

The phenomenon of beats.

A rapid oscillation with a (comparatively) slowly varying periodic amplitude exhibits the phenomenon of beats. For example, if two horns not exactly attuned to one another simultaneously play their middle C, one at ω0/(2π)=258 Hz and the other at ω/(2π)=254 Hz, then one hears a beat—an audible variation in the amplitude of the combined sound—with a frequency of

(ω0ω)/22π=2582542=2 (Hz).

Resonance

Looking at Eq. (6), we see that the amplitude A of xp is large when the natural and external frequencies ω0 and ω are approximately equal. It is sometimes useful to rewrite Eq. (5) in the form

(11)A=F0kmω2=F0/k1(ω/ω0)2=±ρF0k,

where F0/k is the static displacement of a spring with constant k due to a constant force F0, and the amplification factor ρ is defined to be

(12)ρ=1|1(ω/ω0)2|.

It is clear that ρ+ as ωω0. This is the phenomenon of resonance—the increase without bound (as ωω0) in the amplitude of oscillations of an undamped system with natural frequency ω0 in response to an external force with frequency ωω0.

We have been assuming that ωω0. What sort of catastrophe should one expect if ω and ω0 are precisely equal? Then Eq. (4), upon division of each term by m, becomes

(13)x+ω02x=F0mcos ω0t.

Because cosω0t is a term of the complementary function, the method of undetermined coefficients calls for us to try

xp(t)=t(A cos ω0t+B sin ω0t).

We substitute this in Eq. (13) and thereby find that A=0 and B=F0/(2mω0). Hence the particular solution is

(14)xp(t)=F02mω0t sin ω0t,

The graph of xp(t) in Fig. 5.6.4 (in which m=1, F0=100, and ω0=50) shows vividly how the amplitude of the oscillation theoretically would increase without bound in this case of pure resonance, ω=ω0. We may interpret this phenomenon as reinforcement of the natural vibrations of the system by externally impressed vibrations at the same frequency.

FIGURE 5.6.4.

The phenomenon of resonance.

Example 3

Cart with rotating flywheel Suppose that m=5 kg and k=500 N/m in the cart with the flywheel of Fig. 5.6.1. Then the natural frequency is ω0=k/m=10 rad/s; that is, 10/(2π)1.59 Hz. We would therefore expect oscillations of very large amplitude to occur if the flywheel revolves at about (1.59)(60)95 revolutions per minute (rpm).

In practice, a mechanical system with very little damping can be destroyed by resonance vibrations. A spectacular example can occur when a column of soldiers marches in step over a bridge. Any complicated structure such as a bridge has many natural frequencies of vibration. If the frequency of the soldiers’ cadence is approximately equal to one of the natural frequencies of the structure, then—just as in our simple example of a mass on a spring—resonance will occur. Indeed, the resulting resonance vibrations can be of such large amplitude that the bridge will collapse. This has actually happened—for example, the collapse of Broughton Bridge near Manchester, England, in 1831—and it is the reason for the now-standard practice of breaking cadence when crossing a bridge. Resonance may have been involved in the 1981 Kansas City disaster in which a hotel balcony (called a skywalk) collapsed with dancers on it. The collapse of a building in an earthquake is sometimes due to resonance vibrations caused by the ground oscillating at one of the natural frequencies of the structure; this happened to many buildings in the Mexico City earthquake of September 19, 1985. On occasion an airplane has crashed because of resonant wing oscillations caused by vibrations of the engines. It is reported that for some of the first commercial jet aircraft, the natural frequency of the vertical vibrations of the airplane during turbulence was almost exactly that of the mass–spring system consisting of the pilot’s head (mass) and spine (spring). Resonance occurred, causing pilots to have difficulty in reading the instruments. Large modern commercial jets have different natural frequencies, so that this resonance problem no longer occurs.

Modeling Mechanical Systems

The avoidance of destructive resonance vibrations is an ever-present consideration in the design of mechanical structures and systems of all types. Often the most important step in determining the natural frequency of vibration of a system is the formulation of its differential equation. In addition to Newton’s law F=ma, the principle of conservation of energy is sometimes useful for this purpose (as in the derivation of the pendulum equation in Section 5.4). The following kinetic and potential energy formulas are often useful.

  1. Kinetic energy: T=12m v2 for translation of a mass m with velocity v;

  2. Kinetic energy: T=12I ω2 for rotation of a body of a moment of inertia I with angular velocity ω;

  3. Potential energy: V=12k x2 for a spring with constant k stretched or compressed a distance x;

  4. Potential energy: V=mgh for the gravitational potential energy of a mass m at height h above the reference level (the level at which V=0), provided that g may be regarded as essentially constant.

Example 4

Rolling disk Find the natural frequency of a mass m on a spring with constant k if, instead of sliding without friction, it is a uniform disk of radius a that rolls without slipping, as shown in Fig 5.6.5.

FIGURE 5.6.5.

The rolling disk.

Solution

With the preceding notation, the principle of conservation of energy gives

12mv2+12Iω2+12kx2=E

where E is a constant (the total mechanical energy of the system). We note that v=aω and recall that I=ma2/2 for a uniform circular disk. Then we may simplify the last equation to

34mv2+12kx2=E.

Because the right-hand side of this equation is constant, differentiation with respect to t (with v=x and v=x) now gives

34mxx+kxx=0.

We divide each term by 32mx to obtain

x+2k3mx=0.

Thus the natural frequency of horizontal back-and-forth oscillation of our rolling disk is 2k/3m, which is 2/30.8165 times the familiar natural frequency k/m of a mass on a spring that is sliding without friction rather than rolling without sliding. It is interesting (and perhaps surprising) that this natural frequency does not depend on the radius of the disk. It could be either a dime or a large disk with a radius of one meter (but of the same mass).

Example 5

Car suspension Suppose that a car oscillates vertically as if it were a mass m=800 kg on a single spring (with constant k=7×104 N/m), attached to a single dashpot (with constant c=3000 N·s/m). Suppose that this car with the dashpot disconnected is driven along a washboard road surface with an amplitude of 5 cm and a wavelength of L=10 m (Fig. 5.6.6). At what car speed will resonance vibrations occur?

FIGURE 5.6.6.

The washboard road surface of Example 5.

FIGURE 5.6.7.

The “unicycle model” of a car.

Solution

We think of the car as a unicycle, as pictured in Fig. 5.6.7. Let x(t) denote the upward displacement of the mass m from its equilibrium position; we ignore the force of gravity, because it merely displaces the equilibrium position as in Problem 9 of Section 5.4. We write the equation of the road surface as

(15)y=a cos 2πsL(a=0.05 m, L=10 m).

When the car is in motion, the spring is stretched by the amount xy, so Newton’s second law, F=ma, gives

mx=k(xy);

that is,

(16)mx=+k x=ky.

If the velocity of the car is v, then s=vt in Eq. (15), so Eq. (16) takes the form

(16′)mx=+k x=kacos2πvtL.

This is the differential equation that governs the vertical oscillations of the car. In comparing it with Eq. (4), we see that we have forced oscillations with circular frequency ω=2πv/L. Resonance will occur when ω=ω0=k/m. We use our numerical data to find the speed of the car at resonance:

v=L2πkm=102π7×10480014.89 (m/s);

that is, about 33.3 mi/h (using the conversion factor of 2.237 mi/h per m/s).

Damped Forced Oscillations

In real physical systems there is always some damping, from frictional effects if nothing else. The complementary function xc of the equation

(17)mx+c x+k x=F0 cos ωt

is given by Eq. (19), (20), or (21) of Section 5.4, depending on whether c>ccr=4km, c=ccr, or c<ccr. The specific form is not important here. What is important is that, in any case, these formulas show that xc(t)0 as t+. Thus xc is a transient solution of Eq. (17)—one that dies out with the passage of time, leaving only the particular solution xp.

The method of undetermined coefficients indicates that we should substitute

x(t)=A cos ωt+B sin ωt

in Eq. (17). When we do this, collect terms, and equate coefficients of cos ωt and sin ωt, we obtain the two equations

(18)(kmω2)A+cωB=F0,cωA+(kmω2)B=0

that we solve without difficulty for

(19)A=(kmω2)F0(kmω2)2+(cω)2,B=cωF0(kmω2)2+(cω)2.

If we write

A cos ωt+B sin ωt=C(cos ωt cos α+sin ωt sin α)=C cos(ωtα)

as usual, we see that the resulting steady periodic oscillation

(20)xp(t)=C cos(ωtα)

has amplitude

(21)C=A2+B2=F0(kmω2)2+(cω)2.

Now (19) implies that sin α=B/C>0, so it follows that the phase angle α lies in the first or second quadrant. Thus

(22)tan α=BA=cωkmω2with0<α<π,

so

α={tan1cωk=mω2if k>mω2,π+tan1cωk=mω2if k<mω2

(whereas α=π/2 if k=mω2).

Note that if c>0, then the “forced amplitude”—defined as a function C(ω) by (21)—always remains finite, in contrast with the case of resonance in the undamped case when the forcing frequency ω equals the critical frequency ω0=k/m. But the forced amplitude may attain a maximum for some value of ω, in which case we speak of practical resonance. To see if and when practical resonance occurs, we need only graph C as a function of ω and look for a global maximum. It can be shown (Problem 27) that C is a steadily decreasing function of ω if c2km. But if c<2km, then the amplitude of C attains a maximum value—and so practical resonance occurs—at some value of ω less than ω0, and then approaches zero as ω+. It follows that an underdamped system typically will undergo forced oscillations whose amplitude is

Example 6

Practical resonance Find the transient motion and steady periodic oscillations of a damped mass-and-spring system with m=1, c=2, and k=26 under the influence of an external force F(t)=82 cos 4t with x(0)=6 and x(0)=0. Also investigate the possibility of practical resonance for this system.

Solution

The resulting motion x(t)=xtr(t)+xsp(t) of the mass satisfies the initial value problem

(23)x+2x+26x=82 cos 4t;x(0)=6,x(0)=0.

Instead of applying the general formulas derived earlier in this section, it is better in a concrete problem to work it directly. The roots of the characteristic equation

r2+2r+26=(r+1)2+25=0

are r=1±5i, so the complementary function is

xc(t)=et(c1 cos 5t+c2 sin 5t).

When we substitute the trial solution

x(t)=A cos 4t+B sin 4t

in the given equation, collect like terms, and equate coefficients of cos 4t and sin 4t, we get the equations

10A+8B=82,8A+10B=0

with solution A=5, B=4. Hence the general solution of the equation in (23) is

x(t)=et(c1 cos 5t+c2 sin 5t)+5 cos 4t+4 sin 4t.

At this point we impose the initial conditions x(0)=6, x(0)=0 and find that c1=1 and c2=3. Therefore, the transient motion and the steady periodic oscillation of the mass are given by

xtr(t)=et(cos 5t3 sin 5t)

and

xsp(t)=5 cos 4t+4 sin 4t=41(541 cos 4t+441 sin 4t)=41 cos(4tα)

where α=tan1(45)0.6747.

Figure 5.6.8 shows graphs of the solution x(t)=xtr(t)+xsp(t) of the initial value problem

(24)x+2x+26x=82 cos 4t,x(0)=x0,x(0)=0

for the different values x0=20,10, 0, 10, and 20 of the initial position. Here we see clearly what it means for the transient solution xtr(t) to “die out with the passage of time,” leaving only the steady periodic motion xsp(t). Indeed, because xtr(t)0 exponentially, within a very few cycles the full solution x(t) and the steady periodic solution xsp(t) are virtually indistinguishable (whatever the initial position x0).

To investigate the possibility of practical resonance in the given system, we substitute the values m=1, c=2, and k=26 in (21) and find that the forced amplitude at frequency ω is

C(ω)=8267648ω2+ω4.

The graph of C(ω) is shown in Fig. 5.6.9. The maximum amplitude occurs when

C(ω)=41(4ω396ω)(67648ω2+ω4)3/2=164(ω224)(67648ω2+ω4)3/2=0.

Thus practical resonance occurs when the external frequency is ω=24 (a bit less than the mass-and-spring’s undamped critical frequency of ω0=k/m=26).

FIGURE 5.6.8.

Solutions of the initial value problem in (24) with x0=20, 10, 0, 10, and 20.

FIGURE 5.6.9.

Plot of amplitude C versus external frequency ω.

5.6 Problems

In Problems 1 through 6, express the solution of the given initial value problem as a sum of two oscillations as in Eq. (8). Throughout, primes denote derivatives with respect to time t. In Problems 1–4, graph the solution function x(t) in such a way that you can identify and label (as in Fig. 5.6.2) its period.

  1. x+9x=10 cos 2t; x(0)=x(0)=0

  2. x+4x=5 sin 3t; x(0)=x(0)=0

  3. x+100x=225 cos 5t+300 sin 5t; x(0)=375, x(0)=0

  4. x+25x=90 cos 4t; x(0)=0, x(0)=90

  5. mx+kx=F0cos ωt with ωω0; x(0)=x0, x(0)=0

  6. mx+kx=F0cos ωt with ω=ω0; x(0)=0, x(0)=v0

In each of Problems 7 through 10, find the steady periodic solution xsp(t)=Ccos(ωtα) of the given equation mx+cx+kx=F(t) with periodic forcing function F(t) of frequency ω. Then graph xsp(t) together with (for comparison) the adjusted forcing function F1(t)=F(t)/mω.

  1. x+4x+4x=10 cos 3t

  2. x+3x+5x=4 cos 5t

  3. 2x+2x+x=3 sin 10t

  4. x+3x+3x=8 cos 10t+6 sin 10t

In each of Problems 11 through 14, find and plot both the steady periodic solution xsp(t)=Ccos(ωtα) of the given differential equation and the actual solution x(t)=xsp(t)+xtr(t) that satisfies the given initial conditions.

  1. x+4x+5x=10 cos 3t; x(0)=x(0)=0

  2. x+6x+13x=10 sin 5t; x(0)=x(0)=0

  3. x+2x+26x=600 cos 10t; x(0)=10, x(0)=0

  4. x+8x+25x=200cos t+520sin t; x(0)=30, x(0)=10

Each of Problems 15 through 18 gives the parameters for a forced mass–spring–dashpot system with equation mx+cx+kx=F0cos ωt. Investigate the possibility of practical resonance of this system. In particular, find the amplitude C(ω) of steady periodic forced oscillations with frequency ω. Sketch the graph of C(ω) and find the practical resonance frequency ω (if any).

  1. m=1, c=2, k=2, F0=2

  2. m=1, c=4, k=5, F0=10

  3. m=1, c=6, k=45, F0=50

  4. m=1, c=10, k=650, F0=100

  5. A mass weighing 100 lb (mass m=3.125 slugs in fps units) is attached to the end of a spring that is stretched 1 in. by a force of 100 lb. A force F0cos ωt acts on the mass. At what frequency (in hertz) will resonance oscillations occur? Neglect damping.

  6. A front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight W=mg (with g=9.8 m/s2) of the machine depresses the pad exactly 0.5 cm. When its rotor spins at ω radians per second, the rotor exerts a vertical force F0cos ωt newtons on the machine. At what speed (in revolutions per minute) will resonance vibrations occur? Neglect friction.

  7. Pendulum-spring system Figure 5.6.10 shows a mass m on the end of a pendulum (of length L) also attached to a horizontal spring (with constant k). Assume small oscillations of m so that the spring remains essentially horizontal and neglect damping. Find the natural circular frequency ω0 of motion of the mass in terms of L, k, m, and the gravitational constant g.

  8. Pulley-spring system A mass m hangs on the end of a cord around a pulley of radius a and moment of inertia I, as shown in Fig. 5.6.11. The rim of the pulley is attached to a spring (with constant k). Assume small oscillations so that the spring remains essentially horizontal and neglect friction. Find the natural circular frequency of the system in terms of m, a, k, I, and g.

    FIGURE 5.6.10.

    The pendulum-and-spring system of Problem 21.

    FIGURE 5.6.11.

    The mass–spring–pulley system of Problem 22.

  9. Earthquake A building consists of two floors. The first floor is attached rigidly to the ground, and the second floor is of mass m=1000 slugs (fps units) and weighs 16 tons (32,000 lb). The elastic frame of the building behaves as a spring that resists horizontal displacements of the second floor; it requires a horizontal force of 5 tons to displace the second floor a distance of 1 ft. Assume that in an earthquake the ground oscillates horizontally with amplitude A0 and circular frequency ω, resulting in an external horizontal force F(t)=mA0ω2 sin ωt on the second floor. (a) What is the natural frequency (in hertz) of oscillations of the second floor? (b) If the ground undergoes one oscillation every 2.25 s with an amplitude of 3 in., what is the amplitude of the resulting forced oscillations of the second floor?

  10. A mass on a spring without damping is acted on by the external force F(t)=F0cos3 ωt. Show that there are two values of ω for which resonance occurs, and find both.

  11. Derive the steady periodic solution of

    mx+cx+k x=F0 sin ωt.

    In particular, show that it is what one would expect—the same as the formula in (20) with the same values of C and ω, except with sin(ωtα) in place of cos(ωtα).

  12. Given the differential equation

    mx+cx+k x=E0 cos ωt+F0 sin ωt.

    —with both cosine and sine forcing terms—derive the steady periodic solution

    xsp(t)=E02+F02(kmω2)2+(cω)2cos(ωtαβ),

    where α is defined in Eq. (22) and β=tan1(F0/E0). (Suggestion: Add the steady periodic solutions separately corresponding to E0cos ωt and F0sin ωt (see Problem 25).)

  13. According to Eq. (21), the amplitude of forced steady periodic oscillations for the system mx+cx+kx=F0cos ωt is given by

    C(ω)=F0(kmω2)2+(cω)2.

    (a) If cccr/2, where ccr=4km, show that C steadily decreases as ω increases. (b) If c<ccr/2, show that C attains a maximum value (practical resonance) when

    ω=ωm=kmc22m2<ω0=km.
  14. As indicated by the cart-with-flywheel example discussed in this section, an unbalanced rotating machine part typically results in a force having amplitude proportional to the square of the frequency ω. (a) Show that the amplitude of the steady periodic solution of the differential equation

    mx+cx+kx=mAω2 cos ωt

    (with a forcing term similar to that in Eq. (17)) is given by

    C(ω)=mAω2(kmω2)2+(cω)2.

    (b) Suppose that c2<2mk. Show that the maximum amplitude occurs at the frequency ωm given by

    ωm=km(2mk2mkc2).

    Thus the resonance frequency in this case is larger (in contrast with the result of Problem 27) than the natural frequency ω0=k/m. (Suggestion: Maximize the square of C.)

Automobile Vibrations

Problems 29 and 30 deal further with the car of Example 5. Its upward displacement function satisfies the equation mx+cx+kx=cy+ky when the shock absorber is connected (so that c>0). With y=asin ωt for the road surface, this differential equation becomes

mx+cx+kx=E0 cos ωt+F0 sin ωt

where E0=cωa and F0=ka.

  1. Apply the result of Problem 26 to show that the amplitude C of the resulting steady periodic oscillation for the car is given by

    C=ak2+(cω)2(kmω2)2+(cω)2.

    Because ω=2πv/L when the car is moving with velocity v, this gives C as a function of v.

  2. Figure 5.6.12 shows the graph of the amplitude function C(ω) using the numerical data given in Example 5 (including c=3000 N·s/m). It indicates that, as the car accelerates gradually from rest, it initially oscillates with amplitude slightly over 5 cm. Maximum resonance vibrations with amplitude about 14 cm occur around 32 mi/h, but then subside to more tolerable levels at high speeds. Verify these graphically based conclusions by analyzing the function C(ω). In particular, find the practical resonance frequency and the corresponding amplitude.

    FIGURE 5.6.12.

    Amplitude of vibrations of the car on a washboard surface.

5.6 Application Forced Vibrations

Here we investigate forced vibrations of the mass–spring–dashpot system with equation

(1)mx+cx+kx=F(t).

To simplify the notation, let’s take m=p2, c=2p, and k=p2q2+1, where p>0 and q>0. Then the complementary function of Eq. (1) is

(2)xc(t)=et/p(c1 cos qt+c2 sin qt).

We will take p=5, q=3, and thus investigate the transient and steady periodic solutions corresponding to

(3)25x+10x+226x=F(t),x(0)=0,x(0)=0

with several illustrative possibilities for the external force F(t). For your personal investigations to carry out similarly, you might select integers p and q with 6p9 and 2q5.

Investigation 1: With periodic external force F(t)=901 cos 3t, the Matlab commands

x = dsolve(′25*D2x+10*Dx+226*x=901*cos(3*t)′,
     ′x(0)=0, Dx(0)=0′);
x = simple(x);
syms t, xsp = cos(3*t) + 30*sin(3*t);
ezplot(x, [0  6*pi]),hold on
ezplot(xsp, [0  6*pi])

produce the plot shown in Fig. 5.6.13. We see the (transient plus steady periodic) solution

x(t)=cos 3t+30 sin 3t+et/5(cos 3t45115sin 3t)

rapidly “building up” to the steady periodic oscillation xsp(t)=cos 3t+30 sin 3t..

FIGURE 5.6.13.

The solution x(t)=xtr(t)+xsp(t) and the steady periodic solution x(t)=xsp(t) with periodic external force F(t)=901 cos 3t.

Investigation 2: With damped oscillatory external force

F(t)=900et/5 cos 3t,

we have duplication with the complementary function in (2). The Maple commands

de2 := 25*diff(x(t),t,t)+10*diff(x(t),t)+226*x(t) =
       900*exp(-t/5)*cos(3*t);
dsolve({de2,x(0)=0,D(x)(0)=0}, x(t));
x := simplify(combine(rhs(%),trig));
C := 6*t*exp(-t/5);
plot({x,C,-C},t=0..8*Pi);

produce the plot shown in Fig. 5.6.14. We see the solution

x(t)=6tet/5 sin 3t

oscillating up and down between the envelope curves x=±6tet/5. (Note the factor of t that signals a resonance situation.)

FIGURE 5.6.14.

The solution x(t)=6tet/5 and the envelope curves x(t)=±6tet/5 with damped oscillatory force F(t)=900t/5cos 3t.

Investigation 3: With damped oscillatory external force

F(t)=2700tet/5 cos 3t,

we have a still more complicated resonance situation. The Mathematica commands

de3 = 25 x″[t] + 10 x′[t] + 226 x[t] ==
      2700 t Exp[-t/5] Cos[3t]
soln = DSolve[{de3, x[0] == 0, x′[0] == 0}, x[t], t]
x = First[x[t]  /.   soln]
amp = Exp[-t/5] Sqrt[(3t)^2 + (9t^2 - 1)^2]
Plot[{x, amp, -amp}, {t, 0, 10 Pi}]

produce the plot shown in Fig. 5.6.15. We see the solution

x(t)=et/5 [3t cos 3t+(9t21)sin 3t]

oscillating up and down between the envelope curves

x=±et/5(3t)2+(9t21)2.

FIGURE 5.6.15.

The solution x(t)=et/5 [3t cos 3t+(9t21)sin 3t] and the envelope curves x=±et/5(3t)2+(9t21)2 with external force F(t)=2700tet/5 cos 3t.

6 Eigenvalues and Eigenvectors

6.1 Introduction to Eigenvalues

Given a square matrix A, let us pose the following question: Does there exist a nonzero vector v such that the result Av of multiplying v by the matrix A is simply a scalar multiple of v? Thus we ask whether or not there exist a nonzero vector v and a scalar λ such that

(1)Av=λv.

Section 6.3 and subsequent chapters include interesting applications in which this question arises. The following definition provides the terminology that we use in discussing Eq. (1).

The prefix eigen is a German word that (in some contexts) may be translated as proper, so eigenvalues are sometimes called proper values. Eigenvalues and eigenvectors are also called characteristic values and characteristic vectors in some books.

Example 1

Consider the 2×2 matrix

A=[5622].

If v=(2,1)=[21]T, then

Av=[5622][21]=[42]=2v.

Thus, v=[21]T is an eigenvector of A that is associated with the eigenvalue λ=2. If v=[32]T, then

Av=[5622][32]=[32]=1v,

so v=[32]T is an eigenvector of A associated with the eigenvalue λ=1. In summary, we see that the scalars λ1=2 and λ2=1 are both eigenvalues of the matrix A they correspond to the eigenvectors v1=[21]T and v2=[32]T, respectively.

Eigenvalues and eigenvectors have a simple geometric interpretation. Suppose that λ is an eigenvalue of the matrix A and has the associated eigenvector v, so that Av=λv. Then the length |Av| of the vector Av is ±λ|v|, depending on the sign of λ. Thus, if λ>0, then multiplication of v by the matrix A expands or contracts the vector v while preserving its direction; if λ<0, then multiplication of v by A reverses the direction of v (see Fig. 6.1.1).

FIGURE 6.1.1.

(a) A positive eigenvalue; (b) a negative eigenvalue.

Remark 1

If v=0, then the equation Av=λv holds for every scalar λ and hence is of no significance. This is why only nonzero vectors qualify as eigenvectors in the definition.

Remark 2

Let λ and v be an eigenvalue and associated eigenvector of the matrix A. If k is any nonzero scalar and u=kv, then

Au=A(kv)=k(Av)=k(λv)=λ(kv)=λu,

so u=kv is also an eigenvector associated with λ. Thus, any nonzero scalar multiple of an eigenvector is also an eigenvector and is associated with the same eigenvalue. In Example 1, for instance, u1=3v1=[63]T is an eigenvector associated with λ1=2, and u2=4v2=[128]T is an eigenvector associated with λ2=1.

The Characteristic Equation

We now attack the problem of finding the eigenvalues and eigenvectors of a given n×n square matrix A. According to the definition, the nonzero vector v is an eigenvector of A associated with the eigenvalue λ exactly when

Av=λv=λIv;

that is, when

(2)(AλI)v=0.

For a fixed value of λ, Eq. (2) is a homogeneous linear system of n equations in the n components of v. By Theorem 2 in Section 3.6 and Theorem 7 in Section 3.5, this system has a nontrivial solution v0 if and only if the determinant

det(AλI)=|AλI|

of its coefficient matrix is zero. The equation |AλI|=0 is called the characteristic equation of the square matrix A, and we have proved that there exists an eigenvector v associated with λ if and only if λ satisfies this equation.

Now let us see just what sort of equation the characteristic equation in (3) is. Note that the matrix

(4)AλI=[a11λa12a1na21a22λa2nan1an2annλ]

is obtained simply by subtracting λ from each diagonal element of A. If we think of expanding the determinant by cofactors, we see that |AλI| is a polynomial in the variable λ, and that the highest power of λ comes from the product of the diagonal elements of the matrix in (4). Therefore, the characteristic equation of the n×n matrix A takes the form

(5)(1)nλn+cn1λn1++c1λ+c0=0,

an nth-degree polynomial equation in λ.

According to the fundamental theorem of algebra, every nth-degree polynomial equation in one variable has n solutions (counting multiple solutions), but some of them can be complex. Hence, we can say that an n×n matrix A always has n eigenvalues, though they might not be distinct and might not all be real. In this chapter we confine our attention mainly to real eigenvalues, but in Chapter 7 we will see important applications of complex eigenvalues to the solution of linear systems of differential equations.

Solving the characteristic equation is almost always easier said than done. In the examples that follow and in the problems, we have chosen matrices for which the characteristic polynomial |AλI| readily factors to reveal the eigenvalues.

Example 2

Find the eigenvalues and associated eigenvectors of the matrix

A=[5724].

Solution

Here

(6)AλI=[5λ724λ],

so the characteristic equation of A is

0=|AλI|=|5λ724λ|=(5λ)(4λ)(2)(7)=(λ5)(λ+4)+14=λ2λ6;

that is, (λ+2)(λ3)=0. Thus, the matrix A has the two eigenvalues 2 and 3. To distinguish them, we write λ1=2 and λ2=3. To find the associated eigenvectors, we must separately substitute each eigenvalue in (6) and then solve the resulting system (AλI)v=0.

Case 1: λ1=2. With v=[xy]T, the system (AλI)v=0 is

[7722][xy]=[00].

Each of the two scalar equations here is a multiple of the equation x+y=0, and any nontrivial solution v=[xy]T of this equation is a nonzero multiple of [11]T. Hence, to within a constant multiple, the only eigenvector associated with λ1=2 is v1=[11]T.

Case 2: λ2=3. With v=[xy]T, the system (AλI)v=0 is

[2727][xy]=[00].

Again, we have only a single equation, 2x+7y=0, and any nontrivial solution of this equation will suffice. The choice y=2 yields x=7, so (to within a constant multiple) the only eigenvector associated with λ2=3 is v2=[72]T.

Finally, note that it is not enough to say simply that the given matrix A has eigenvalues 2 and 3 and has eigenvectors [11]T and [72]T. To give complete information, we must say which eigenvector is associated with each eigenvalue.

If the elements of the matrix A are all real, then so are the coefficients of its characteristic equation |AλI|=0. In this event, it follows that the complex eigenvalues (if any) of A occur only in conjugate pairs. The next example illustrates the possibility of imaginary or complex eigenvalues and eigenvectors.

Example 3

The characteristic equation of the matrix

A=[0820]

is

|AλI|=[λ82λ]=λ2+16=0.

Hence the matrix A has the complex conjugate eigenvalues λ=±4i.

Case 1: λ1=4i. With v=[xy]T, the system (AλI)v=0 is

[4i824i][xy]=[00].

The first equation here, 4ix+8y=0, is 2i times the second one and, obviously, is satisfied by x=2i, y=1. Thus v1=[2i1]T is a complex eigenvector associated with the complex eigenvalue λ1=4i.

Case 2: λ2=+4i. In this case, you may verify similarly that the conjugate v2=[2i1]T of v1 is an eigenvector associated with the conjugate λ2 of the eigenvalue λ1. Thus the complex conjugate eigenvalues 4i of the matrix A correspond to the complex conjugate eigenvectors [±2i1]T (taking either both upper signs or both lower signs).

Example 4

The 2×2 identity matrix I has characteristic equation

|1λ001λ|=(1λ)2=0,

so I has the single eigenvalue λ=1. The equation (I1λ)v=0 is

[0000][xy]=[00],

so every nonzero vector v=[xy]T is an eigenvector of I. In particular, the single eigenvalue λ=1 corresponds to the two linearly independent eigenvectors v1=[10]T and v2=[01]T.

Example 5

The characteristic equation of the matrix

A=[2302]

is (2λ)2=0, so A has the single eigenvalue λ=2. The equation (A2I)v=0 is

[0300][xy]=[00].

Thus x is arbitrary, but y=0, so the eigenvalue λ=2 corresponds (to within a constant multiple) to the single eigenvector v=[10]T.

Examples 25 illustrate the four possibilities for a 2×2 matrix A. It can have either

  • two distinct real eigenvalues, each corresponding to a single eigenvector;

  • a single real eigenvalue corresponding to a single eigenvector;

  • a single real eigenvalue corresponding to two linearly independent eigenvectors; or

  • two complex conjugate eigenvalues corresponding to complex conjugate eigenvectors.

The characteristic equation of a 3×3 matrix is, by Eq. (5), of the form

(7)λ3+c2λ2+c1λ+c0=0.

Now the value of the continuous function λ3++c0 on the left-hand side here approaches + as λ and as λ+. Hence every such cubic equation has at least one real solution, so every 3×3 matrix has (in contrast with 2×2 matrices) at least one real eigenvalue. A 3×3 matrix may have one, two, or three distinct eigenvalues, and a single eigenvalue of a 3×3 matrix may correspond to one, two, or three linearly independent eigenvectors. The remaining two examples of this section illustrate some of the possibilities. The next example shows also that, whereas the zero vector 0 cannot be an eigenvector of a matrix, there is nothing to prevent λ=0 from being an eigenvalue.

Example 6

Find the eigenvalues and associated eigenvectors of the matrix

A=[30046216155].

Solution

The matrix AλI is

(8)AλI=[3λ0046λ216155λ].

Upon expansion of the determinant along its first row, we find that

|AλI|=(3λ)[(6λ)(5λ)+30]=(3λ)(λ2λ)=λ(λ1)(3λ).

Hence the characteristic equation |AλI|=0 yields the three eigenvalues λ1=0, λ2=1, and λ3=3. To find the associated eigenvectors, we must solve the system (AλI)v=0 separately for each of these three eigenvalues.

Case 1: λ1=0. We write v=[xyz]T and substitute λ=0 in the coefficient matrix in (8) to obtain the system

[30046216155][xyz]=[000].

From the first of the three equations here, 3x=0, we see that x=0. Then each of the remaining two equations is a multiple of the equation 3y+z=0. The choice y=1 yields z=3. Thus the eigenvector v1=[013]T is associated with λ1=0.

Case 2: λ2=1. Substitution of λ=1 in the coefficient matrix in (8) yields the system

[20045216156][xyz]=[000]

for v=[xyz]T. The first equation 2x=0 implies that x=0. Then the third equation is a multiple of the second equation, 5y+2z=0. The choice of y=2 yields z=5, so the eigenvector v2=[025]T is associated with λ2=1.

Case 3: λ3=3. Substitution of λ=3 in the coefficient matrix in (8) yields the system

[00043216158][xyz]=[000].

In this case, the first equation yields no information, but the result of adding 4 times the second equation to the third equation is 3y=0, so y=0. Consequently, the second and third equations are both multiples of the equation 2xz=0. The choice x=1 yields z=2, so the eigenvector v3=[102]T is associated with λ3=3.

In summary, we have found the eigenvectors v1=[013]T, v2=[025]T, and v3=[102]T associated with the distinct eigenvalues λ1=0, λ2=1, and λ3=3, respectively.

Remark

Substitution of λ=0 in the characteristic equation |AλI|=0 yields |A|=0. Therefore, λ=0 is an eigenvalue of the matrix A if and only if A is singular: |A|=0.

Eigenspaces

Let λ be a fixed eigenvalue of the n×n matrix A. Then the set of all eigenvectors associated with A is the set of all nonzero solution vectors of the system

(9)(AλI)v=0.

The solution space of this system is called the eigenspace of A associated with the eigenvalue λ. This subspace of Rn consists of all eigenvectors associated with λ together with the zero vector. In Example 6 we found (to within a constant multiple) only a single eigenvector associated with each eigenvalue λ; in this case, the eigenspace of λ is 1-dimensional. In the case of an eigenspace of higher dimension, we generally want to find a basis for the solution space of Eq. (9).

Example 7

Find bases for the eigenspaces of the matrix

A=[421201223].

Solution

Here we have

(10)AλI=[4λ212λ1223λ].

We expand along the first row to obtain

|AλI|=(4λ)(λ23λ+2)(2)(42λ)+(1)(4+2λ)=λ3+7λ216λ+12.

Thus, to find the eigenvalues we need to solve the cubic equation

(11)λ37λ2+16λ12=0.

We look (hopefully) for integer solutions. The factor theorem of algebra implies that if the polynomial equation

λn+cn1λn1++c1λ+c0=0

with integer coefficients and leading coefficient 1 has an integer solution, then that integer is a divisor of the constant c0. In the case of the cubic equation in (11), the possibilities for such a solution are ±1, ±2, ±3, ±4, ±6, and ±12. We substitute these numbers successively in (11) and thereby find that +1 and 1 are not solutions but that λ=+2 is a solution. Hence λ2 is a factor of the cubic polynomial in (11). Next, the long division

λ25λ+6λ2λ37λ2+16λ12λ32λ2_5λ2+16λ125λ2+10λ_6λ126λ12_0

shows that

λ37λ2+16λ12=(λ2)(λ25λ+6)=(λ2)(λ2)(λ3)=(λ2)2(λ3).

Thus we see finally that the given matrix A has the repeated eigenvalue λ=2 and the eigenvalue λ=3.

Case 1: λ=2. The system (AλI)v=0 is

[221221221][xyz]=[000],

which reduces to the single equation 2x2y+z=0. This equation obviously has a 2-dimensional solution space. With y=1 and z=0, we get x=1 and, hence, obtain the basis eigenvector v1=[110]T. With y=0 and z=2, we get x=1 and, hence, the basis eigenvector v2=[102]T. The 2-dimensional eigenspace of A associated with the repeated eigenvalue λ=2 has basis {v1,v2}.

Case 2: λ=3. The system (AλI)v=0 is

[121231220][xyz]=[000].

The last equation here implies that x=y, and then each of the first two equations yields x=y=z. It follows that the eigenspace of A associated with λ=3 is 1-dimensional and has v3=[111]T as a basis eigenvector.

Remark

The typical higher-degree polynomial is not so easy to factor as the one in Example 7. Hence a numerical technique such as Newton’s method is often needed to solve the characteristic equation. Moreover, for an n×n matrix with n greater than about 4, the amount of labor required to find the characteristic equation by expanding the determinant |AλI| is generally prohibitive; because of the presence of the variable λ, row and column elimination methods do not work as they do with numerical determinants. Consequently, specialized techniques, beyond the scope of the present discussion, are often required to find the eigenvalues and eigenvectors of the large matrices that occur in many applications. Problems 40 and 41 at the end of this section outline a numerical technique that sometimes is useful with matrices of moderate size.

6.1 Problems

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.

  1. [4211]

  2. [5634]

  3. [8631]

  4. [4321]

  5. [10965]

  6. [6431]

  7. [10864]

  8. [761210]

  9. [81021]

  10. [91020]

  11. [19102110]

  12. [131566]

  13. [200221263]

  14. [5004422126]

  15. [220221223]

  16. [101231660]

  17. [352020021]

  18. [10068212153]

  19. [362010001]

  20. [10047210154]

  21. [431211002]

  22. [563673662]

  23. [1222022200320004]

  24. [1040014000300003]

  25. [1010011000200002]

  26. [4003020000106005]

Find the complex conjugate eigenvalues and corresponding eigenvectors of the matrices given in Problems 27 through 32.

  1. A=[0110]

  2. A=[0660]

  3. A=[03120]

  4. A=[012120]

  5. A=[02460]

  6. A=[04360]

  7. Suppose that λ is an eigenvalue of the matrix A with associated eigenvector v and that n is a positive integer. Show that λn is an eigenvalue of An with associated eigen-vector v.

  8. Show that λ is an eigenvalue of the invertible matrix A if and only if λ1 is an eigenvalue of A1. Are the associated eigenvectors the same?

    1. Suppose that A is a square matrix. Use the characteristic equation to show that A and AT have the same eigenvalues.

    2. Give an example of a 2×2 matrix A such that A and AT do not have the same eigenvectors.

  9. Show that the eigenvalues of a triangular n×n matrix are its diagonal elements.

  10. Suppose that the characteristic equation |AλI|=0 is written as a polynomial equation [Eq. (5)]. Show that the constant term is c0=detA. Suggestion: Substitute an appropriate value for λ.

Problems 38 through 42 introduce the trace of a square matrix and explore its connections with the determinant and the characteristic polynomial of the matrix.

  1. If A=[aij] is an n×n matrix, then the trace Tr A of A is defined to be

    Tr A=a11+a22++ann,

    the sum of the diagonal elements of A. It can be proved that the coefficient of λn1 in Eq. (5) is cn1=(1)n1(Tr A). Show explicitly that this is true in the case of a 2×2 matrix.

  2. Suppose that the n×n matrix A has n (real) eigenvalues λ1,λ2,,λn. Assuming the general result stated in Problem 38, prove that

    λ1+λ2++λn=Tr A=a11+a22++ann.
  3. According to the results stated in Problems 37 and 38, the characteristic polynomial

    p(λ)=|AλI|

    of a 3×3 matrix A is given by

    p(λ)=λ3+(Tr A)λ2+c1λ+(detA).

    The remaining coefficient c1 can be found by substituting λ=1 and then calculating the two determinants |A| and p(1)=|AI|. Use this method to find the characteristic equation, eigenvalues, and associated eigenvectors of the matrix

    A=[326747714137156].
  4. According to the results stated in Problems 37 and 38, the characteristic polynomial

    p(λ)=|AλI|

    of a 4×4 matrix A is given by

    p(λ)=λ4(Tr A)λ3+c2λ2+c1λ+(detA).

    The remaining coefficients c1 and c2 can be found by substituting λ=±1 and calculating the three determinants |A|, p(1)=|AI|, and p(1)=|A+I|. Use this method to find the characteristic equation, eigenvalues, and associated eigenvectors of the matrix

    A=[22988107142100810299315].
  5. Combine ideas from Problems 37–40 to show explicitly that

    1. If A is a 2×2 matrix then det(AλI)=(λ)2+Tr (A)(λ)+det(A).

    2. If A is a 3×3 matrix then det(AλI)=(λ)3+Tr (A)(λ)2+c1(λ)det(A), where c1 is the sum of the minors of the diagonal elements of A. (Recall from Section 3.6 the distinction between minors and cofactors.)

6.2 Diagonalization of Matrices

Given an n×n matrix A, we may ask how many linearly independent eigenvectors the matrix A has. In Section 6.1, we saw several examples (with n=2 and n=3) in which the n×n matrix A has n linearly independent eigenvectors—the largest possible number. By contrast, in Example 5 of Section 6.1, we saw that the 2×2 matrix

A=[2302]

has the single eigenvalue λ=2 corresponding to the single eigenvector v=[10]T.

Something very nice happens when the n×n matrix A does have n linearly independent eigenvectors. Suppose that the eigenvalues λ1,λ2,,λn (not necessarily distinct) of A correspond to the n linearly independent eigenvectors v1,v2,,vn, respectively. Let

(1)P=[|||v1v2vn|||]

be the n×n matrix having these eigenvectors as its column vectors. Then

AP=A[|||v1v2vn|||]=[|||Av2Av2Avn|||]

and hence

(2)AP=[|||λ1v1λ2v2λnvn|||],

because Avj=λjvj for each j=1,2,,n. Thus the product matrix AP has column vectors λ1v1,λ2v2,,λnvn.

Now consider the diagonal matrix

(3)D=[λ1000λ2000λn],

whose diagonal elements are the eigenvalues corresponding (in the same order) to the eigenvectors forming the columns of P. Then

(4)PD=[|||v1v2vn|||][λ1000λ2000λn]=[|||λ1v2λ2v2λnvn|||],

because the product of the ith row of P and the jth column of D is simply the product of λj and the ith component of vj.

Finally, upon comparing the results in (2) and (4), we see that

(5)AP=PD.

But the matrix P is invertible, because its n column vectors are linearly independent. So we may multiply on the right by P1 to obtain

(6)A=PDP1.

Equation (6) expresses the n×n matrix A having n linearly independent eigenvectors in terms of the eigenvector matrix P and the diagonal eigenvalue matrix D. It can be rewritten as D=P1AP, but the form in (6) is the one that should be memorized.

Example 1

In Example 1 of Section 6.1 we saw that the matrix

A=[5622]

has eigenvalues λ1=2 and λ2=1 corresponding to the linearly independent eigenvectors v1=[21]T and v2=[32]T, respectively. Then

P=[2312],D=[2001],andP1=[2312].

So

PDP1=[2312][2001][2312]=[4322][2312]=[5622]=A,

in accord with Eq. (6).

Similarity and Diagonalization

The following definition embodies the precise relationship in (6) between the original matrix A and the diagonal matrix D.

Note that this relationship between A and B is symmetric, for if B=P1AP, then A=Q1BQ for some invertible matrix Q—just take Q=P1.

An n×n matrix A is called diagonalizable if it is similar to a diagonal matrix D; that is, there exist a diagonal matrix D and an invertible matrix P such that A=PDP1, and so

(8)P1AP=D.

The process of finding the diagonalizing matrix P and the diagonal matrix D in (8) is called diagonalization of the matrix A. In Example 1 we showed that the matrices

A=[5622]andD=[2001]

are similar, and hence that the 2×2 matrix A is diagonalizable.

Now we ask under what conditions a given square matrix is diagonalizable. In deriving Eq. (6), we showed that if the n×n matrix A has n linearly independent eigenvectors, then A is diagonalizable. The converse of this statement is also true.

Remark

It is important to remember not only the fact that an n×n matrix A having n linearly independent eigenvectors is diagonalizable, but also the specific diagonalization A=PDP1 in Eq. (6), where the matrix P has the n eigenvectors as its columns, and the corresponding eigenvalues are the diagonal elements of the diagonal matrix D.

Example 2

In Example 5 of Section 6.1 we saw that the matrix

A=[2302]

has only one eigenvalue, λ=2, and that (to within a constant multiple) only the single eigenvector v=[10]T is associated with this eigenvalue. Thus the 2×2 matrix A does not have n=2 linearly independent eigenvectors. Hence Theorem 1 implies that A is not diagonalizable.

Example 3

In Example 6 of Section 6.1 we saw that the matrix

A=[30046216155]

has the following eigenvalues and associated eigenvectors:

λ1=3:v1=[102]Tλ1=1:v2=[025]Tλ1=0:v3=[013]T.

It is obvious (why?) that the three eigenvectors v1, v2, v3 are linearly independent, so Theorem 1 implies that the 3×3 matrix A is diagonalizable. In particular, the inverse of the eigenvector matrix

P=[v1v2v3]=[100021253]

is

P1=[100231452],

and the diagonal eigenvalue matrix is

D=[λ1000λ2000λ3]=[300010000].

Therefore, Eq. (6) in the form P1AP=D yields the diagonalization

P1AP=[100231452] [30046216155] [100021253]=[300010000]

of the matrix A.

The following theorem tells us that any set of eigenvectors associated with distinct eigenvalues (as in Example 3) is automatically linearly independent.

If the n×n matrix A has n distinct eigenvalues, then by Theorem 2 the n associated eigenvectors are linearly independent, so Theorem 1 implies that the matrix A is diagonalizable. Thus we have the sufficient condition for diagonalizability given in Theorem 3.

In general, however, an n×n matrix A can be expected to have fewer than n distinct eigenvalues λ1,λ2,,λk. If k<n, then we may attempt to diagonalize A by carrying out the following procedure.

  1. Step 1. Find a basis Si for the eigenspace associated with each eigenvalue λi.

  2. Step 2. Form the union S of the bases S1,S2,,Sk. According to Theorem 4 in this section, the set S of eigenvectors of A is linearly independent.

  3. Step 3. If S contains n eigenvectors v1,v2,,vn, then the matrix

    P=[v1v2vn]

    diagonalizes A: that is, P1AP=D, where the diagonal elements of D are the eigenvalues (repeated as necessary) corresponding to the n eigenvectors v1, v2, , vn.

If the set S—obtained by “merging” the bases for all the eigenspaces of A—contains fewer than n eigenvectors, then it can be proved that the matrix A is not diagonalizable.

Example 4

In Example 7 of Section 6.1, we saw that the matrix

A=[421201223]

has only two distinct eigenvalues, λ1=2 and λ2=3. We found that the eigenvalue λ1=2 corresponds to a 2-dimensional eigenspace with basis vectors v1=[110]T and v2=[102]T, and that λ2=3 corresponds to a 1-dimensional eigenspace with basis vector v3=[111]T. By Theorem 4 (or by explicit verification), these three eigenvectors are linearly independent, so Theorem 1 implies that the 3×3 matrix A is diagonalizable. The diagonalizing matrix

P=[v1v2v3]=[111101021]

has inverse matrix

P1=[231110221],

so we obtain the diagonalization

P1AP=[231110221] [421201223] [111101021]=[200020003]=D

of the matrix A.

6.2 Problems

In Problems 1 through 28, determine whether or not the given matrix A is diagonalizable. If it is, find a diagonalizing matrix P and a diagonal matrix D such that P1AP=D.

  1. [5421]

  2. [6644]

  3. [5320]

  4. [5432]

  5. [9865]

  6. [106127]

  7. [61023]

  8. [111568]

  9. [1413]

  10. [3111]

  11. [5191]

  12. [1191613]

  13. [130020002]

  14. [221221221]

  15. [331221001]

  16. [320010441]

  17. [783673222]

  18. [652432223]

  19. [111241441]

  20. [20061126150]

  21. [010120111]

  22. [221120571]

  23. [241351111]

  24. [321101112]

  25. [1020012000100001]

  26. [1001010100110002]

  27. [1100011000110002]

  28. [1101011100210002]

  29. Prove: If the matrices A and B are similar and the matrices B and C are similar, then the matrices A and C are similar.

  30. Suppose that the matrices A and B are similar and that n is a positive integer. Prove that the matrices An and Bn are similar.

  31. Suppose that the invertible matrices A and B are similar. Prove that their inverses A1 and B1 are also similar.

  32. Show that if the n×n matrices A and B are similar, then they have the same characteristic equation and therefore have the same eigenvalues.

  33. Suppose that the n×n matrices A and B are similar and that each has n real eigenvalues. Show that detA=detB and that Tr A=Tr B. See Problems 38 and 39 in Section 6.1.

  34. Consider the 2×2 matrix

    A=[abcd]

    and let Δ=(ad)2+4bc. Then show that

    1. A is diagonalizable if Δ>0;

    2. A is not diagonalizable if Δ<0;

    3. If Δ=0, then A may be diagonalizable or it may not be.

  35. Let A be a 3×3 matrix with three distinct eigenvalues. Tell how to construct six different invertible matrices P1,P2,,P6 and six different diagonal matrices D1,D2,,D6 such that PiDi(Pi)1=A for each i=1,2,,6.

  36. Prove: If the diagonalizable matrices A and B have the same eigenvalues (with the same multiplicities), then A and B are similar.

  37. Given: The diagonalizable matrix A. Show that the eigenvalues of A2 are the squares of the eigenvalues of A but that A and A2 have the same eigenvectors.

  38. Suppose that the n×n matrix A has n linearly independent eigenvectors associated with a single eigenvalue λ. Show that A is a diagonal matrix.

  39. Let λi be an eigenvalue of the n×n matrix A, and assume that the characteristic equation of A has only real solutions. The algebraic multiplicity of λi is the largest positive integer p(i) such that (λλi)p(i) is a factor of the characteristic polynomial |AλI|. The geometric multiplicity of λi is the dimension q(i) of the eigenspace associated with λi. It can be shown that p(i)q(i) for every eigenvalue λi. Taking this as already established, prove that the given matrix A is diagonalizable if and only if the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.

6.3 Applications Involving Powers of Matrices

In this section we discuss applications that depend on an ability to compute the matrix Ak for large values of k, given the n×n matrix A. If A is diagonalizable, then Ak can be found directly by a method that avoids the labor of calculating the powers A2,A3,A4, by successive matrix multiplications.

Recall from Section 6.2 that, if the n×n matrix A has n linearly independent eigenvectors v1,v2,,vn associated with the eigenvalues λ1,λ2,,λn (not necessarily distinct), then

(1)A=PDP1,

where

P=[|||v1v2vn|||]andD=[λ1000λ2000λn].

Note that (1) yields

A2=(PDP1)(PDP1)=PD(P1P)DP1=PD2P1

because P1P=I. More generally, for each positive integer k,

(2)Ak=(PDP1)k=(PDP1)(PDP1)(PDP1)(PDP1)=PD(P1P)D(P1P)(PDP1);Ak=PDkP1.

But the kth power Dk of the diagonal matrix D is easily computed:

(3)Dk=[λ1k000λ2k000λnk].

Consequently the formula in (2) gives a quick and effective way to calculate any desired power of the diagonalizable matrix A, once its eigenvalues and associated eigenvectors—and hence the matrices P and D—have been found.

Example 1

Find A5 if

A=[421201223].

Solution

In Example 7 of Section 6.1, we found that the 3×3 matrix A has the eigenvalue λ1=3 with associated eigenvector v1=[111]T, and the repeated eigenvalue λ2=2 with associated eigenvectors v2=[110]T and v3=[102]T. Therefore, A=PDP1 with

P=[111110102]andD=[300020002].

We first calculate

P1=[221231110]andD5=[2430003200032].

Then the formula in (3) yields

A5=[111110102][2430003200032][221231110]=[2433232243320243064][221231110]=[454422211422390211422422243].

Transition Matrices

We want to apply this method of computing Ak to the analysis of a certain type of physical system that can be described by means of the following kind of mathematical model. Suppose that the sequence

(4)x0,x1,x2,,xk,

of n-vectors is defined by its initial vector x0 and an n×n transition matrix A in the following manner:

(5)xk+1=Axkfor each k0.

We envision a physical system—such as a population with n specified subpopulations—that evolve through a sequence of successive states described by the vectors in (4). Then our goal is to calculate the kth state vector xk. But using (5) repeatedly, we find that

x1=Ax0,x2=Ax1=A2x0,x3=Ax2=A3x0,

and in general that

(6)xk=Akx0.

Thus our task amounts to calculating the kth power Ak of the transition matrix A.

Example 2

Urban/suburban population Consider a metropolitan area with a constant total population of 1 million individuals. This area consists of a city and its suburbs, and we want to analyze the changing urban and suburban populations. Let Ck denote the city population and Sk the suburban population after k years. Suppose that each year 15% of the people in the city move to the suburbs, whereas 10% of the people in the suburbs move to the city. Then it follows that

(7)Ck+1=0.85Ck+0.10SkSk+1=0.15Ck+0.90Sk

for each k0. (For instance, next year’s city population Ck+1 will equal 85% of this year’s city population Ck plus 10% of this year’s suburban population Sk.) Thus the metropolitan area’s population vector xk=[CkSk]T satisfies

(8)xk+1=Axkand hencexk=Akx0

(for each k0) with transition matrix

A=[0.850.100.150.90].

The characteristic equation of the transition matrix A is

(1720λ))(910λ)(320)(110)=0;(1720λ)(910λ)3=0;200λ2350λ+150=0;4λ27λ+3=0;(λ1)(4λ3)=0.

Thus the eigenvalues of A are λ1=1 and λ2=0.75. For λ1=1, the system (AλI)v=0 is

[0.150.100.150.10][xy]=[00],

so an associated eigenvector is v1=[23]T. For λ2=0.75, the system (AλI)v=0 is

[0.100.100.150.15][xy]=[00],

so an associated eigenvector is v2=[11]T. It follows that A=PDP1, where

P=[2131],D=[10034],andP1=15[1132].

Now suppose that our goal is to determine the long-term distribution of population between the city and its suburbs. Note first that (34)k is “negligible” when k is sufficiently large; for instance, (34)400.00001. It follows that if k40, then the formula Ak=PDkP1 yields

(9)Ak=[2131][100(34)k](15)[1132]15[2131][1000][1132]=15[2030][1132]=15[2233].

Hence it follows that, if k is sufficiently large, then

xk=Akx015[2233][C0S0]=(C0+S0)[0.40.6]=[0.40.6],

because C0+S0=1 (million), the constant total population of the metropolitan area. Thus, our analysis shows that, irrespective of the initial distribution of population between the city and its suburbs, the long-term distribution consists of 40% in the city and 60% in the suburbs.

Remark

The result in Example 2—that the long-term situation is independent of the initial situation—is characteristic of a general class of common problems. Note that the transition matrix A in (8) has the property that the sum of the elements in each column is 1. An n×n matrix with nonnegative entries having this property is called a stochastic matrix. It can be proved that, if A is a stochastic matrix having only positive entries, then λ1=1 is one eigenvalue of A and |λi|<1 for the others. (See Problems 39 and 40.) Moreover, as k, the matrix Ak approaches the constant matrix

[v1v1v1],

each of whose identical columns is the eigenvector of A associated with λ1=1 that has the sum of its elements equal to 1. The 2×2 stochastic matrix A of Example 1 illustrates this general result, with λ1=1, λ2=34, and v1=(25,35).

Predator-Prey Models

Next, we consider a predator-prey population consisting of the foxes and rabbits living in a certain forest. Initially, there are F0 foxes and R0 rabbits; after k months, there are Fk foxes and Rk rabbits. We assume that the transition from each month to the next is described by the equations

(10)Fk+1=0.4Fk+0.3RkRk+1=rFk+1.2Rk,

where the constant r>0 is the “capture rate” representing the average number of rabbits consumed monthly by each fox. Thus

(11)xk+1=Axkand hencexk=Akx0,

where

(12)xk=[FkRk]andA=[0.40.3r1.2].

The term 0.4Fk in the first equation in (10) indicates that, without rabbits to eat, only 40% of the foxes would survive each month, while the term 0.3Rk represents the growth in the fox population due to the available food supply of rabbits. The term 1.2Rk in the second equation indicates that, in the absence of any foxes, the rabbit population would increase by 20% each month. We want to investigate the long-term behavior of the fox and rabbit populations for different values of the capture rate r of rabbits by foxes.

The characteristic equation of the transition matrix A in (12) is

(0.4λ)(1.2λ)+(0.3)r=0;(410λ)(1210λ)+30r=0;100λ2160λ+(48+30r)=0.

The quadratic formula then yields the equation

(13)λ=1200[160±(160)2(400)(48+30r)]=110(8±1630r),

which gives the eigenvalues of A in terms of the capture rate r. Examples 3, 4, and 5 illustrate three possibilities (for different values of r) for what may happen to the fox and rabbit populations as k increases:

Example 3

Stable limiting population If r=0.4, then Eq. (13) gives the eigenvalues λ1=1 and λ2=0.6. For λ1=1, the system (AλI)v=0 is

[0.60.30.40.6] [xy]=[00],

so an associated eigenvector is v1=[12]T. For λ2=0.6, the system (AλI)v=0 is

[0.20.30.40.6] [xy]=[00],

so an associated eigenvector is v2=[32]T. It follows that A=PDP1, where

P=[1322],D=[1000.6],andP1=14[2321].

We are now ready to calculate Ak. If k is sufficiently large that (0.6)k0—for instance, (0.6)250.000003—then the formula Ak=PDkP1 yields

Ak=[1322][100(0.6)k](14)[2321]14[1322][1000][2321]=14[1020][2321]=14[2346].

Hence it follows that if k is sufficiently large, then

xk=Akx0=14[2346] [F0R0]=14[3R02F06R04F0]

—that is,

(14)[FkRk]=α[12],whereα=14(3R02F0).

Assuming that the initial populations are such that α>0 (that is, 3R0>2F0), (14) implies that, as k increases, the fox and rabbit populations approach a stable situation in which there are twice as many rabbits as foxes. For instance, if F0=R0=100, then when k is sufficiently large, there will be 25 foxes and 50 rabbits.

Example 4

Mutual extinction If r=0.5, then Eq. (13) gives the eigenvalues λ1=0.9 and λ2=0.7. For λ1=0.9, the system (AλI)v=0 is

[0.50.30.50.3] [xy]=[00],

so an associated eigenvector is v1=[35]T. For λ2=0.7, the system (AλI)v=0 is

[0.30.30.50.5] [xy]=[00],

so an associated eigenvector is v2=[11]T. It follows that A=PDP1, with

P=[3151],D=[0.9000.7],andP1=12[1153].

Now both (0.9)k and (0.7)k approach 0 as k increases without bound (k+). Hence if k is sufficiently large, then the formula Ak=PDkP1 yields

Ak=[3151][(0.9)k00(0.7)k](12)[1153]12[3151][0000][1153]=[0000],

so

(15)[FkRk]=Ak[F0R0][0000][F0R0]=[00].

Thus Fk and Rk both approach zero as k+, so both the foxes and the rabbits die out—mutual extinction occurs.

Example 5

Population explosion If r=0.325, then Eq. (13) gives the eigenvalues λ1=1.05 and λ2=0.55. For λ1=1.05, the system (AλI)v=0 is

[0.6500.300.3250.15][xy]=[00].

Each equation is a multiple of 13x+6y=0, so an associated eigenvector is v1=[613]T. For λ2=0.55, the system (AλI)v=0 is

[0.1500.300.3250.65][xy]=[00],

so an associated eigenvector is v2=[21]T. It follows that A=PDP1 with

P=[62131],D=[1.05000.55],andP1=120[12136].

Note that (0.55)k approaches zero but that (1.05)k increases without bound as k+. It follows that if k is sufficiently large, then the formula Ak=PDkP1 yields

Ak=[62131][(1.05)k00(0.55)k](120)[12136]120[62131][(1.05)k000][12136]=120[(6)(1.05)k0(13)(1.05)k0][12136],

and therefore

(16)Ak120(1.05)k[6121326].

Hence, if k is sufficiently large, then

xk=Akx0120(1.05)k[6121326][F0R0]=120(1.05)k[6F012R013F026R0],

and so

(17)[FkRk](1.05)kγ[613],whereγ=120(2R0F0).

If γ>0 (that is, if 2R0>F0), then the factor (1.05)k in (17) implies that the fox and rabbit populations both increase at a rate of 5% per month, and thus each increases without bound as k+. Moreover, when k is sufficiently large, the two populations maintain a constant ratio of 6 foxes for every 13 rabbits. It is also of interest to note that the monthly “population multiplier” is the larger eigenvalue λ1=1.05 and that the limiting ratio of populations is determined by the associated eigenvector v1=[613]T.

In summary, let us compare the results in Examples 3, 4, and 5. The critical capture rate r=0.4 of Example 3 represents a monthly consumption of 0.4 rabbits per fox, resulting in stable limiting populations of both species. But if the foxes are greedier and consume more than 0.4 rabbits per fox monthly, then the result is extinction of both species (as in Example 4). If the rabbits become more skilled at evading foxes, so that less than 0.4 rabbits per fox are consumed each month, then both populations grow without bound, as in Example 5.

The Cayley-Hamilton Theorem

One of the more notable and important theorems in advanced linear algebra says that every matrix satisfies its own characteristic equation (as proved for the 2×2 case in Problem 29 of Section 3.4).

Example 6

In Example 7 of Section 6.1, we found that the matrix

A=[421201223]

has the characteristic polynomial

p(λ)=λ3+7λ216λ+12,

so

(19)A3+7A216A+12I=0,

by the Cayley-Hamilton theorem. Because

A2=[14105106510109],

it follows from (19) that

A3=7A216A+12I=7[14105106510109]16[421201223]+12[100010001]=[463819383019383827].

Multiplication by A now gives

A4=7A316A2+12A=7(7A216A+12I)16A2+12A=33A2100A+84I=33[14105106510109]100[421201223]+84[100010001]=[146130651301146513013081].

Thus we can use the Cayley-Hamilton theorem to express (positive integral) powers of the 3×3 matrix A in terms of A and A2. We also can use the Cayley-Hamilton theorem to find the inverse matrix A1. If we multiply Eq. (19) by A1, we can solve the resulting equation A2+7A16I+12A1=0:

A1=112(A27A+16I)=112[14105106510109]712[421201223]+1612[100010001]=16[121251222].

6.3 Problems

In Problems 1 through 10, a matrix A is given. Use the method of Example 1 to compute A5.

  1. [3210]

  2. [5634]

  3. [6644]

  4. [4321]

  5. [5432]

  6. [61023]

  7. [130020002]

  8. [121010022]

  9. [131020002]

  10. [431211002]

Find A10 for each matrix A given in Problems 11 through 14.

  1. [10065221156]

  2. [116220114001]

  3. [111221441]

  4. [553221443]

  1. 15–24. Use the Cayley-Hamilton theorem (as in Example 6 ) to find A1, A3, and A4 for each matrix A given in Problems 5-14 (respectively).

In Problems 25 through 30, a city-suburban population transition matrix A (as in Example 2) is given. Find the resulting long-term distribution of a constant total population between the city and its suburbs.

  1. A=[0.90.10.10.9]

  2. A=[0.850.050.150.95]

  3. A=[0.750.150.250.85]

  4. A=[0.80.10.20.9]

  5. A=[0.90.050.10.95]

  6. A=[0.80.150.20.85]

Predator-Prey

Problems 31 through 33 deal with a fox-rabbit population as in Examples 3 through 5, except with the transition matrix

A=[0.60.5r1.2]

in place of the one used in the text.

  1. If r=0.16, show that in the long term the populations of foxes and rabbits are stable, with 5 foxes for each 4 rabbits.

  2. If r=0.175, show that in the long term the populations of foxes and rabbits both die out.

  3. If r=0.135, show that in the long term the fox and rabbit populations both increase at the rate of 5% per month, maintaining a constant ratio of 10 foxes for each 9 rabbits.

  4. Suppose that the 2×2 matrix A has eigenvalues λ1=1 and λ2=1 with eigenvectors v1=(3,4) and v2=(5,7), respectively. Find the matrix A and the powers A99 and A100.

  5. Suppose that |λ|=1 for each eigenvalue λ of the diagonalizable matrix A. Show that An=I for every even positive integer n.

  6. Suppose that

    A=[0110].

    Show that A2n=I and that A2n+1=A for every positive integer n.

  7. Suppose that

    A=[0110].

    Show that A4n=I, A4n+1=A, A4n+2=I, and A4n+3=A for every positive integer n.

  8. The matrix

    A=[1101]

    is not diagonalizable. (Why not?) Write A=I+B. Show that B2=0 and thence that An=[1n01].

  9. Consider the stochastic matrix

    A=[p1q1pq],

    where 0<p<1 and 0<q<1. Show that the eigenvalues of A are λ1=1 and λ2=p+q1, so that |λ2|<1.

  10. Suppose that A is an n×n stochastic matrix-the sum of the elements of each column vector is 1. If v=(1,1,,1), show that ATv=v. Why does it follow that λ=1 is an eigenvalue of A?

  11. In his Liber abaci (Book of Calculation) published in 1202, Leonardo Fibonacci* asked the following question: How many pairs of rabbits are produced from a single original pair in one year, if every month each pair begets a new pair, which is similarly productive beginning in the second succeeding month? The answer is provided by the Fibonacci sequence

    1,1,2,3,5,8,13,21,34,55,

    in which each term is the sum of its two immediate predecessors. That is, the sequence is defined recursively as follows:

    s0=1=s1,sn+1=sn+sn1for n1.

    Then sn is the number of rabbit pairs present after n months. Note that if

    xn=[sn+1sn]andA=[1110],

    then

    xn=Axn1and soxn=Anx0,

    where x0=(1,1).

    1. Show that A has eigenvalues λ1=12(1+5) and λ2=12(15) with associated eigenvectors v1=(1+5,2) and v2=(15,2), respectively.

    2. Compute

      [sn+1sn]=Anx0=PDnP1[11]

      to derive the amazing formula

      sn=15[(1+52)n+1(152)n+1].

      Thus after 1 year the number of rabbit pairs is

      s12=15[(1+52)13(152)13]=233.

      Show that there are 75,025 rabbit pairs after 2 years and over 2.5 trillion rabbit pairs after 5 years.

7 Linear Systems of Differential Equations

7.1 First-Order Systems and Applications

In Chapters 1 and 5 we discussed methods for solving an ordinary differential equation that involves only one dependent variable. Many applications, however, require the use of two or more dependent variables, each a function of a single independent variable (typically time). Such a problem leads naturally to a system of simultaneous ordinary differential equations. We will usually denote the independent variable by t and the dependent variables (the unknown functions of t) by x1, x2, x3, , or by x, y, z, . Primes will indicate derivatives with respect to t.

We will restrict our attention to systems in which the number of equations is the same as the number of dependent variables (unknown functions). For instance, a system of two first-order equations in the dependent variables x and y has the general form

(1)f(t, x, y, x, y)=0,g(t, x, y, x, y)=0,

where the functions f and g are given. A solution of this system is a pair x(t), y(t) of functions of t that satisfy both equations identically over some interval of values of t.

For an example of a second-order system, consider a particle of mass m that moves in space under the influence of a force field F that depends on time t, the position (x(t), y(t), z(t)) of the particle, and its velocity (x(t), y(t), z(t)). Writing Newton’s law ma=F componentwise, we get the system

(2)mx=F1(t, x, y, z, x, y, z),my=F2(t, x, y, z, x, y, z),mz=F3(t, x, y, z, x, y, z)

of three second-order equations with independent variable t and dependent variables x, y, z; the three right-hand side functions F1, F2, F3 are the components of the vector-valued function F.

Initial Applications

Examples 1 and 2 further illustrate how systems of differential equations arise naturally in scientific problems.

Example 1

Dual mass-spring system Consider the system of two masses and two springs shown in Fig. 7.1.1, with a given external force f(t) acting on the right-hand mass m2. We denote by x(t) the displacement (to the right) of the mass m1 from its static equilibrium position [when the system is motionless and in equilibrium and f(t)=0] and by y(t) the displacement of the mass m2 from its static position. Thus the two springs are neither stretched nor compressed when x and y are zero.

FIGURE 7.1.1.

The mass-and-spring system of Example 1.

In the configuration in Fig. 7.1.1, the first spring is stretched x units and the second by yx units. We apply Newton’s law of motion to the two “free body diagrams” shown in Fig. 7.1.2; we thereby obtain the system

(3)m1x=k1x+k2(yx),m2y=k2(yx)+f(t)

FIGURE 7.1.2.

The free body diagrams for the system of Example 1.

of differential equations that the position functions x(t) and y(t) must satisfy. For instance, if m1=2, m2=1, k1=4, k2=2, and f(t)=40 sin 3t in appropriate physical units, then the system in (3) reduces to

(4)2x=6x+2y,y=2x2y+40 sin 3t.

Example 2

Dual brine tanks Consider two brine tanks connected as shown in Fig. 7.1.3. Tank 1 contains x(t) pounds of salt in 100 gal of brine and tank 2 contains y(t) pounds of salt in 200 gal of brine. The brine in each tank is kept uniform by stirring, and brine is pumped from each tank to the other at the rates indicated in Fig. 7.1.3. In addition, fresh water flows into tank 1 at 20 gal/min, and the brine in tank 2 flows out at 20 gal/min (so the total volume of brine in the two tanks remains constant). The salt concentrations in the two tanks are x/100 pounds per gallon and y/200 pounds per gallon, respectively. When we compute the rates of change of the amount of salt in the two tanks, we therefore get the system of differential equations that x(t) and y(t) must satisfy:

x=30x100+10y200=310x+120y,y=30x10010y20020y200=310x3020y

FIGURE 7.1.3.

The two brine tanks of Example 2.

—that is,

(5)20x=6x+y,20y=6x3y.

First-Order Systems

Consider a system of differential equations that can be solved for the highest-order derivatives of the dependent variables that appear, as explicit functions of t and lower-order derivatives of the dependent variables. For instance, in the case of a system of two second-order equations, our assumption is that it can be written in the form

(6)x1=f1(t, x1, x2, x1, x2),x2=f2(t, x1, x2, x1, x2).

It is of both practical and theoretical importance that any such higher-order system can be transformed into an equivalent system of first-order equations.

To describe how such a transformation is accomplished, we consider first the “system” consisting of the single nth-order equation

(7)x(n)=f(t, x, x, , x(n1)).

We introduce the dependent variables x1, x2, , xn defined as follows:

(8)x1=x,x2=x,x3=x,,xn=x(n1).

Note that x1=x=x2, x2=x=x3, and so on. Hence the substitution of (8) in Eq. (7) yields the system

(9)x1=x2,x2=x3,xn1=xn,xn=f(t, x1, x2, xn)

of n first-order equations. Evidently, this system is equivalent to the original nth-order equation in (7), in the sense that x(t) is a solution of Eq. (7) if and only if the functions x1(t), x2(t), , xn(t) defined in (8) satisfy the system of equations in (9).

Example 3

The third-order equation

x(3)+3x+2x5x=sin 2t

is of the form in (7) with

f(t, x, x, x)=5x2x3x+sin 2t.

Hence the substitutions

x1=x,x2=x=x1,x3=x=x2

yield the system

x1=x2,x2=x3,x3=5x12x23x3+sin 2t

of three first-order equations.

It may appear that the first-order system obtained in Example 3 offers little advantage, because we could use the methods of Chapter 5 to solve the original (linear) third-order equation. But suppose that we were confronted with the nonlinear equation

x=x3+(x)3,

to which none of our earlier methods can be applied. The corresponding first-order system is

(10)x1=x2,x2=(x1)3+(x2)3,

and we will see in Section 7.7 that there exist effective numerical techniques for approximating the solution of essentially any first-order system. So, in this case, the transformation to a first-order system is advantageous. From a practical viewpoint, large systems of higher-order differential equations typically are solved numerically with the aid of the computer, and the first step is to transform such a system into a first-order system for which a standard computer program is available.

Example 4

The system

(4)2x=6x+2y,y=2x2y+40 sin 3t

of second-order equations was derived in Example 1. Transform this system into an equivalent first-order system.

Solution

Motivated by the equations in (8), we define

x1=x,x2=x=x1,y1=y,y2=y=y1.

Then the system in (4) yields the system

(11)x1=x2,2x2=6x1+2y1,y1=y2,y2=2x12y1+40 sin 3t

of four first-order equations in the dependent variables x1, x2, y1, and y2.

Simple Two-Dimensional Systems

The linear second-order differential equation

(12)x+px+qx=0

(with constant coefficients and independent variable t) transforms via the substitutions x=y, x=y into the two-dimensional linear system

(13)x=y,y=qxpy.

Conversely, we can solve this system in (13) by solving the familiar single equation in (12).

Example 5

To solve the two-dimensional system

(14)x=2y,y=12x,

we begin with the observation that

x=2y=2(12x)=x.

This gives the single second-order equation x+x=0 having general solution

x(t)=A cos t+B sin t=C cos(tα),

where A=C cos α and B=C sin α. Then

y(t)=12x(t)=12(A sin t+B cos t)=12C sin(tα).

The identity cos2 θ+sin2 θ=1 therefore implies that, for each value of t, the point (x(t), y(t)) lies on the ellipse

x2C2+y2(C/2)2=1

with semiaxes C and C/2. Figure 7.1.4 shows several such ellipses in the xy-plane.

FIGURE 7.1.4.

Direction field and solution curves for the system x=2y, y=12x of Example 5.

A solution (x(t), y(t)) of a two-dimensional system

x=f(t, x, y),y=g(t, x, y)

can be regarded as a parametrization of a solution curve or trajectory of the system in the xy-plane. Thus the trajectories of the system in (14) are the ellipses of Fig. 7.1.4. The choice of an initial point (x(0), y(0)) determines which one of these trajectories a particular solution parametrizes.

The picture showing a system’s trajectories in the xy-plane—its so-called phase plane portrait—fails to reveal precisely how the point (x(t), y(t)) moves along its trajectory. If the functions f and g do not involve the independent variable t, then a direction field—showing typical arrows representing vectors with components (proportional to) the derivatives x=f(x, y) and y=g(x, y)—can be plotted. Because the moving point (x(t), y(t)) has velocity vector (x(t), y(t)), this direction field indicates the point’s direction of motion along its trajectory. For instance, the direction field plotted in Fig. 7.1.4 indicates that each such point moves counterclockwise around its elliptical trajectory. Additional information can be shown in the separate graphs of x(t) and y(t) as functions of t.

FIGURE 7.1.5.

x- and y-solution curves for the initial value problem x=2y,y=12x,x(0)=2,y(0)=0.

Example 5

Continued With initial values x(0)=2, y(0)=0, the general solution in Example 5 yields

x(0)=A=2,y(0)=12B=0.

The resulting particular solution is given by

x(t)=2 cos t,y(t)=sin t.

The graphs of the two functions are shown in Fig. 7.1.5. We see that x(t) initially decreases while y(t) increases. It follows that, as t increases, the solution point (x(t), y(t)) traverses the trajectory 14x2+y2=1 in the counterclockwise direction, as indicated by the direction field vectors in Fig. 7.1.4.

Example 6

To find a general solution of the system

(15)x=y,y=2x+y,

we begin with the observation that

x=y=2x+y=x+2x.

This gives the single linear second-order equation

xx2x=0

having the characteristic equation

r2r2=(r+1)(r2) =0

and the general solution

(16)x(t)=Aet+Be2t.

Therefore,

(17)y(t)=x(t)=Aet+2Be2t.

Typical phase plane trajectories of the system in (15) parametrized by Eqs. (16) and (17) are shown in Fig. 7.1.6. These trajectories may resemble hyperbolas sharing common asymptotes, but Problem 23 shows that their actual form is somewhat more complicated.

FIGURE 7.1.6.

Direction field and solution curves for the system x=y, y=2x+y of Example 6.

Example 7

To solve the initial value problem

(18)x=y,y=(1.01)x(0.2)y,x(0)=0,y(0)=1,

we begin with the observation that

x=y=[(1.01)x(0.2)y]=(1.01)x(0.2)x.

This gives the single linear second-order equation

x+(0.2)x+(1.01)x=0

having the characteristic equation

r2+(0.2)r+1.01=(r+0.1)2+1=0,

characteristic roots 0.1±i, and the general solution

x(t)=et/10(A cos t+B sin t).

Then x(0)=A=0, so

x(t)=Bet/10 sin t,y(t)=x(t)=110Bet/10 sin tBet/10 cos t.

Finally, y(0)=B=1, so the desired solution of the system in (18) is

(19)x(t)=et/10 sin t,y(t)=110et/10(sin t+10 cos t).

These equations parametrize the spiral trajectory in Fig.7.1.7; the trajectory approaches the origin as t+. Figure 7.1.8 shows the x- and y-solution curves given in (19).

FIGURE 7.1.7.

Direction field and solution curve for the system x=y, y=(1.01)x(0.2)y of Example 7.

FIGURE 7.1.8.

x- and y-solution curves for the initial value problem of Example 7.

When we study linear systems in subsequent sections, we will learn why the superficially similar systems in Examples 5 through 7 have the markedly different trajectories shown in Figs. 7.1.4, 7.1.6, and 7.1.7.

Linear Systems

In addition to practical advantages for numerical computation, the general theory of systems and systematic solution techniques are more easily and more concisely described for first-order systems than for higher-order systems. For instance, consider a linear first-order system of the form

(20)x1=p11(t)x1+p12(t)x2++p1n(t)xn+f1(t),x2=p21(t)x1+p22(t)x2++p2n(t)xn+f2(t),xn=pn1(t)x1+pn2(t)x2++pnn(t)xn+fn(t).

We say that this system is homogeneous if the functions f1, f2, , fn are all identically zero; otherwise, it is nonhomogeneous. Thus the linear system in (5) is homogeneous, whereas the linear system in (11) is nonhomogeneous. The system in (10) is nonlinear because the right-hand side of the second equation is not a linear function of the dependent variables x1 and x2.

A solution of the system in (20) is an n-tuple of functions x1(t), x2(t), , xn(t) that (on some interval) identically satisfy each of the equations in (20). We will see that the general theory of a system of n linear first-order equations shares many similarities with the general theory of a single nth-order linear differential equation. Theorem 1 (proved in Appendix A) is analogous to Theorem 2 of Section 5.2. It tells us that if the coefficient functions pij and fj in (20) are continuous, then the system has a unique solution satisfying given initial conditions.

Thus n initial conditions are needed to determine a solution of a system of n linear first-order equations, and we therefore expect a general solution of such a system to involve n arbitrary constants. For instance, we saw in Example 4 that the second-order linear system

2x=6x+2y,y=2x2y+40 sin 3t,

which describes the position functions x(t) and y(t) of Example 1, is equivalent to the system of four first-order linear equations in (11). Hence four initial conditions would be needed to determine the subsequent motions of the two masses in Example 1. Typical initial values would be the initial positions x(0) and y(0) and the initial velocities x(0) and y(0). On the other hand, we found that the amounts x(t) and y(t) of salt in the two tanks of Example 2 are described by the system

20x=6x+y,20y=6x3y

of two first-order linear equations. Hence the two initial values x(0) and y(0) should suffice to determine the solution. Given a higher-order system, we often must transform it into an equivalent first-order system to discover how many initial conditions are needed to determine a unique solution. Theorem 1 tells us that the number of such conditions is precisely the same as the number of equations in the equivalent first-order system.

7.1 Problems

In Problems 1 through 10, transform the given differential equation or system into an equivalent system of first-order differential equations.

  1. x+3x+7x=t2

  2. x(4)+6x3x+x=cos 3t

  3. t2x+tx+(t21)x=0

  4. t3x(3)2t2x+3tx+5x=ln t

  5. x(3)=(x)2+cos x

  6. x5x+4y=0, y+4x5y=0

  7. x=kx(x2+y2)3/2, y=ky(x2+y2)3/2

  8. x+3x+4x2y=0, y+2y3x+y=cos t

  9. x=3xy+2z, y=x+y4z, z=5xyz

  10. x=(1y)x, y=(1x)y

Use the method of Examples 5, 6, and 7 to find general solutions of the systems in Problems 11 through 20. If initial conditions are given, find the corresponding particular solution. For each problem, use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system.

  1. x=y, y=x

  2. x=y, y=x

  3. x=2y, y=2x; x(0)=1, y(0)=0

  4. x=10y, y=10x; x(0)=3, y(0)=4

  5. x=12y, y=8x

  6. x=8y, y=2x

  7. x=y, y=6xy; x(0)=1, y(0)=2

  8. x=y, y=10x7y; x(0)=2, y(0)=7

  9. x=y, y=13x+4y; x(0)=0, y(0)=3

  10. x=y, y=9x+6y

    1. Calculate [x(t)]2+[y(t)]2 to show that the trajectories of the system x=y, y=x of Problem 11 are circles.

    2. Calculate [x(t)]2[y(t)]2 to show that the trajectories of the system x=y, y=x of Problem 12 are hyperbolas.

    1. Beginning with the general solution of the system x=2y, y=2x of Problem 13, calculate x2+y2 to show that the trajectories are circles.

    2. Show similarly that the trajectories of the system x=12y, y=8x of Problem 15 are ellipses with equations of the form 16x2+y2=C2.

  11. First solve Eqs. (16) and (17) for et and e2t in terms of x(t), y(t), and the constants A and B. Then substitute the results in (e2t)(et)2=1 to show that the trajectories of the system x=y, y=2x+y in Example 6 satisfy an equation of the form

    4x33xy2+y3=C(constant).

    Then show that C=0 yields the straight lines y=x and y=2x that are visible in Fig. 7.1.6.

  12. Dual mass-spring system Derive the equations

    m1x1=(k1+k2)x1+k2x2,m2x2=k2x1(k2+k3)x2

    for the displacements (from equilibrium) of the two masses shown in Fig. 7.1.9.

    FIGURE 7.1.9.

    The system of Problem 24.

  13. Oscillating particles Two particles each of mass m are attached to a string under (constant) tension T, as indicated in Fig. 7.1.10. Assume that the particles oscillate vertically (that is, parallel to the y-axis) with amplitudes so small that the sines of the angles shown are accurately approximated by their tangents. Show that the displacements y1 and y2 satisfy the equations

    ky1=2y1+y2,ky2=y12y2,

    where k=mL/T.

    FIGURE 7.1.10.

    The mechanical system of Problem 25.

  14. Fermentation vats Three 100-gal fermentation vats are connected as indicated in Fig. 7.1.11, and the mixtures in each tank are kept uniform by stirring. Denote by xi(t) the amount (in pounds) of alcohol in tank Ti at time t (i=1, 2, 3). Suppose that the mixture circulates between the tanks at the rate of 10 gal/min. Derive the equations

    10x1=x1+x310x2=x1x210x3=x2x3.

    FIGURE 7.1.11.

    The fermentation tanks of Problem 26.

  15. A particle of mass m moves in the plane with coordinates (x(t), y(t)) under the influence of a force that is directed toward the origin and has magnitude k/(x2+y2)—an inverse-square central force field. Show that

    mx=kxr3andmy=kyr3,

    where r=x2+y2.

  16. Suppose that a projectile of mass m moves in a vertical plane in the atmosphere near the surface of the earth under the influence of two forces: a downward gravitational force FG of magnitude mg, and a resistive force FR that is directed opposite to the velocity vector v and has magnitude kv2 (where v=|v| is the speed of the projectile; see Fig. 7.1.12). Show that the equations of motion of the projectile are

    mx=kvx,my=kvymg,

    where v=(x)2+(y)2.

    FIGURE 7.1.12.

    The trajectory of the projectile of Problem 28.

  17. Suppose that a particle with mass m and electrical charge q moves in the xy-plane under the influence of the magnetic field B=Bk (thus a uniform field parallel to the z-axis), so the force on the particle is F=qv×B if its velocity is v. Show that the equations of motion of the particle are

    mx=+qBy,my=qBx.

7.1 Application Gravitation and Kepler’s Laws of Planetary Motion

Around the turn of the seventeenth century, Johannes Kepler analyzed a lifetime of planetary observations by the astronomer Tycho Brahe. Kepler concluded that the motion of the planets around the sun is described by the following three propositions, now known as Kepler’s laws of planetary motion:

  1. The orbit of each planet is an ellipse with the sun at one focus.

  2. The radius vector from the sun to each planet sweeps out area at a constant rate.

  3. The square of the planet’s period of revolution is proportional to the cube of the major semiaxis of its elliptical orbit.

In his Principia Mathematica (1687) Isaac Newton deduced the inverse square law of gravitation from Kepler’s laws. In this application we lead you (in the opposite direction) through a derivation of Kepler’s first two laws from Newton’s law of gravitation. Newton’s explanation of planetary motion in terms of gravitation is a landmark of scientific and intellectual history.

Assume that the sun is located at the origin in the plane of motion of a planet, and write the position vector of the planet in the form

(1)r(t)=(x(t), y(t))=xi+yj,

where i=(1,0) and j=(0,1) denote the unit vectors in the positive x- and y-directions. Then the inverse-square law of gravitation implies (Problem 27) that the acceleration vector r(t) of the planet is given by

(2)r=krr3,

where r=x2+y2 is the distance from the sun to the planet. If the polar coordinates of the planet at time t are (r(t),θ(t)), then the radial and transverse unit vectors shown in Fig. 7.1.13 are given by

(3)ur=i cos θ+j sin θanduθ=i sin θ+j cos θ.

FIGURE 7.1.13.

The radial and transverse unit vectors ur and uθ.

The radial unit vector ur (when located at the planet’s position) always points directly away from the origin, so ur=r/r, and the transverse unit vector uθ is obtained from ur by a 90° counterclockwise rotation.

Step 1: Differentiate the equations in (3) componentwise to show that

(4)durdt=uθdθdtandduθdt=urdθdt.

Step 2: Use the equations in (4) to differentiate the planet’s position vector r=rur and thereby show that its velocity vector is given by

(5)v=drdt=urdrdt+rdθdtuθ.

Step 3: Differentiate again to show that the planet’s acceleration vector a=dv/dt is given by

(6)a=[d2rdt2r(dθdt)2]ur+[1rddt(r2dθdt)]uθ.

Step 4: The radial and transverse components on the right-hand sides in Eqs. (2) and (6) must agree. Equating the transverse components—that is, the coefficients of uθ—we get

(7)1rddt(r2dθdt)=0,

so it follows that

(8)r2dθdt=h,

where h is a constant. Because the polar-coordinate area element—for computation of the area A(t) in Fig. 7.1.14—is given by dA=12r2dθ, Eq. (8) implies that the derivative A(t) is constant, which is a statement of Kepler’s second law.

FIGURE 7.1.14.

Area swept out by the radius vector.

Step 5: Equate radial components in (2) and (6), and then use the result in (8) to show that the planet’s radial coordinate function r(t) satisfies the second-order differential equation

(9)d2rdt2h2r3=kr2.

Step 6: Although the differential equation in (9) is nonlinear, it can be transformed to a linear equation by means of a simple substitution. For this purpose, assume that the orbit can be written in the polar-coordinate form r=r(θ), and first use the chain rule and Eq. (8) to show that if r=1/z, then

drdt=hdzdθ.

Differentiate again, to deduce from Eq. (9) that the function z(θ)=1/r(θ) satisfies the second-order equation

(10)d2zdθ2+z=kh2.

Step 7: Show that the general solution of Eq. (10) is

(11)z(θ)=A cos θ+B sin θ+kh2.

Step 8: Finally, deduce from Eq. (11) that r(θ)=1/z(θ) is given by

(12)r(θ)=L1+ecos(θα),

with e=Ch2/k, Ccos α=A, Csin α=B, and L=h2/k. The polar-coordinate graph of Eq. (12) is a conic section of eccentricity e—an ellipse if 0e<1, a parabola if e=1, and a hyperbola if e>1—with focus at the origin. Planetary with perihelion distance r1=L/(1+e) and aphelion distance r2=L/(1e). orbits are bounded and therefore are ellipses, with eccentricity e<1. As indicated in Fig. 7.1.15, the major axis of the ellipse lies along the radial line θ=α.

FIGURE 7.1.15.

The elliptical orbit

r=L1+ecos(θα)

with perihelion distance

r1=L/(1+e) and aphelion distance

r2=L/(1e).

Step 9: Plot some typical elliptical orbits—as described by (12)—having different eccentricities, sizes, and orientations. In rectangular coordinates you can write

x(t)=r(t)cos t,y(t)=r(t)sin t,0t2π

to plot an elliptical orbit with eccentricity e, semilatus rectum L (Fig. 7.1.15), and rotation angle α. The eccentricity of the earth’s orbit is e0.0167, so close to zero that the orbit looks nearly circular (though with the sun off center), and the eccentricities of the other planetary orbits range from 0.0068 for Venus and 0.0933 for Mars to 0.2056 for Mercury (and 0.2486 for Pluto, no longer classified a planet). But many comets have highly eccentric orbits—Halley’s comet has e0.97 (Fig.7.1.16).

FIGURE 7.1.16.

The shape of the orbit of Halley’s comet.

7.2 Matrices and Linear Systems

A system of differential equations often can be simplified by expressing it as a single differential equation involving a matrix-valued function. A matrix-valued function, or simply matrix function, is a matrix such as

(1)x(t)=[x1(t)x2(t)xn(t)]

or

(2)A(t)=[a11(t)a12(t)a1n(t)a21(t)a22(t)a2n(t)am1(t)am2(t)amn(t)],

in which each entry is a function of t. We say that the matrix function A(t) is continuous (or differentiable) at a point (or on an interval) if each of its elements has the same property. The derivative of a differentiable matrix function is defined by elementwise differentiation; that is,

(3)A(t)=dAdt=[daijdt].

Example 1

If

x(t)=[tt2et]andA(t)=[sint1tcost],

then

dxdt=[12tet]andA(t)=[cost01sint].

The differentiation rules

(4)ddt(A+B)=dAdt+dBdt

and

(5)ddt(AB)=AdBdt+dAdtB

follow readily by elementwise application of the analogous differentiation rules of elementary calculus for real-valued functions. If c is a (constant) real number and C is a constant matrix, then

(6)ddt(cA)=cdAdt,ddt(CA)=CdAdt,andddt(AC)=dAdtC.

Because of the noncommutativity of matrix multiplication, it is important not to reverse the order of the factors in Eqs. (5) and (6).

First-Order Linear Systems

The notation and terminology of matrices and vectors may seem rather elaborate when first encountered, but it is readily assimilated with practice. Our main use for matrix notation will be the simplification of computations with systems of differential equations, especially those computations that would be burdensome in scalar notation.

We discuss here the general system of n first-order linear equations

(7)x1=p11(t)x1+p12(t)x2++p1n(t)xn+f1(t),x2=p12(t)x1+p22(t)x2++p2n(t)xn+f2(t),x3=p13(t)x1+p32(t)x2++p3n(t)xn+f3(t),xn=pn1(t)x1+pn2(t)x2++pnn(t)xn+fn(t).

If we introduce the coefficient matrix

P(t)=[pij(t)]

and the column vectors

x=[xi]andf(t)=[fi(t)],

then—as illustrated in Example 2 below—the system in (7) takes the form of a single matrix equation

(8)dxdt=P(t)x+f(t).

We will see that the general theory of the linear system in (7) closely parallels that of a single nth-order equation. The matrix notation used in Eq. (8) not only emphasizes this analogy, but also saves a great deal of space.

A solution of Eq. (8) on the open interval I is a column vector function x(t)=[xi(t)] such that the component functions of x satisfy the system in (7) identically on I. If the functions pij(t) and fi(t) are all continuous on I, then Theorem 1 of Section 7.1 guarantees the existence on I of a unique solution x(t) satisfying preassigned initial conditions x(a)=b.

Example 2

The first-order system

x1=4x13x2,x2=6x17x2

can be recast as a matrix equation by writing

x=[x1x2]=[4x13x26x17x2]=[4367][x1x2]=[4367]x.

Thus we obtain a matrix equation of the form in (8),

dxdt=P(t)x+f(t)withP(t)=[4   36   7]andf(t)=[00]=0.

To verify that the vector functions

x1(t)=[3e2t2e2t]andx2(t)=[ e5t3e5t]

are both solutions of the matrix differential equation with coefficient matrix P, we need only calculate

Px1=[4367][3e2t2e2t]=[6e2t4e2t]=x1

and

Px2=[4367][e5t3e5t]=[5e5t15e5t]=x2.

To investigate the general nature of the solutions of Eq. (8), we consider first the associated homogeneous equation

(9)dxdt=P(t)x,

which has the form shown in Eq. (8), but with f(t)0. We expect it to have n solutions x1, x2, , xn that are independent in some appropriate sense, and that are such that every solution of Eq. (9) is a linear combination of these n particular solutions. Given n solutions x1, x2, , xn of Eq. (9), let us write

(10)xj(t)=[x1j(t)xij(t)xnj(t)].

Thus xij(t) denotes the ith component of the vector xj(t), so the second subscript refers to the vector function xj(t), whereas the first subscript refers to a component of this function. Theorem 1 is analogous to Theorem 1 of Section 5.2.

Proof: We know that xi=P(t)xi for each i (1in), so it follows immediately that

x=c1x1+c2x2++cnxn=c1P(t)x1+c2P(t)x2++cnP(t)xn=P(t)(c1x1+c2x2++cnxn).

That is, x=P(t)x, as desired. The remarkable simplicity of this proof demonstrates clearly one advantage of matrix notation.

Example 2

Continued If x1 and x2 are the two solutions of

dxdt=[4367]x

discussed in Example 2, then the linear combination

x(t)=c1x1(t)+c2x2(t)=c1[3e2t2e2t]+c2[e5t3e5t]

is also a solution. In scalar form with x=[x1x2]T, this gives the solution

x1(t)=3c1e2t+c2e5t,x2(t)=2c1e2t+3c2e5t.

Independence and General Solutions

Linear independence is defined in the same way for vector-valued functions as for real-valued functions (Section 5.2). The vector-valued functions x1, x2, , xn are linearly dependent on the interval I provided that there exist constants c1, c2, , cn, not all zero, such that

(12)c1x1(t)+c2x2(t)++cnxn(t)=0

for all t in I. Otherwise, they are linearly independent. Equivalently, they are linearly independent provided that no one of them is a linear combination of the others. For instance, the two solutions x1 and x2 of Example 2 are linearly independent because, clearly, neither is a scalar multiple of the other.

Just as in the case of a single nth-order equation, there is a Wronskian determinant that tells us whether or not n given solutions of the homogeneous equation in (9) are linearly dependent. If x1, x2, , xn are such solutions, then their Wronskian is the n×n determinant

(13)W(t)=|x11(t)x12(t)x1n(t)x21(t)x22(t)x2n(t)xn1(t)xn2(t)xnn(t)|,

using the notation in (10) for the components of the solutions. We may write either W(t) or W(x1,x2,,xn). Note that W is the determinant of the matrix that has as its column vectors the solutions x1, x2, , xn. Theorem 2 is analogous to Theorem 3 of Section 5.2. Moreover, its proof is essentially the same, with the definition of W(x1,x2,,xn) in Eq. (13) substituted for the definition of the Wronskian of n solutions of a single nth-order equation. (See Problems 34 through 36.)

Example 3

It is readily verified (as in Example 2) that

x1(t)=[2et2etet],x2(t)=[2e3t0e3t],andx3(t)=[2e5t2e5te5t]

are solutions of the equation

(14)dxdt=[320132013]x.

The Wronskian of these solutions is

W=|2et2e3t2e5t2et02e5tete3te5t|=e9t|222202111|=16e9t,

which is never zero. Hence Theorem 2 implies that the solutions x1, x2, and x3 are linearly independent (on any open interval).

Theorem 3 is analogous to Theorem 4 of Section 5.2. It says that the general solution of the homogeneous n×n system x=P(t)x is a linear combination

(15)x=c1x1+c2x2++cnxn

of any n given linearly independent solutions x1, x2, , xn.

Proof

Let a be a fixed point of I. We first show that there exist numbers c1, c2, , cn such that the solution

(16)y(t)=c1x1(t)+c2x2(t)++cnxn(t)

has the same initial values at t=a as does the given solution x(t); that is, such that

(17)c1x1(a)+c2x2(a)++cnxn(a)=x(a).

Let X(t) be the n×n matrix with column vectors x1, x2, , xn, and let c be the column vector with components c1, c2, , cn. Then Eq. (17) may be written in the form

(18)X(a)c=x(a).

The Wronskian determinant W(a)=|X(a)| is nonzero because the solutions x1, x2, , xn are linearly independent. Hence the matrix X(a) has an inverse matrix X(a)1. Therefore the vector c=X(a)1x(a) satisfies Eq. (18), as desired.

Finally, note that the given solution x(t) and the solution y(t) of Eq. (16)—with the values of ci determined by the equation c=X(a)1x(a)—have the same initial values (at t=a). It follows from the existence-uniqueness theorem of Section 7.1 that x(t)=y(t) for all t in I. This establishes Eq. (15).

Remark Every n×n system x=P(t)x with continuous coefficient matrix does have a set of n linearly independent solutions x1, x2, , xn as in the hypotheses of Theorem 3. It suffices to choose for xj(t) the unique solution such that

xj(a)=[0000100]position j

—that is, the column vector with all elements zero except for a 1 in row j. (In other words, xj(a) is merely the jth column of the identity matrix.) Then

W(x1,x2,,xn)|t=a=|I|0,

so the solutions x1, x2, , xn are linearly independent by Theorem 2. How to actually find these solutions explicitly is another matter—one that we address in Section 7.3 (for the case of constant-coefficient matrices).

Initial Value Problems and Elementary Row Operations

The general solution in Eq. (15) of the homogeneous linear system x=P(t)x can be written in the form

(19)x(t)=X(t)c,

where

(20)X(t)=[x1(t)x2(t)xn(t)]

is the n×n matrix whose column vectors are the linearly independent solutions x1, x2, , xn, and where c=[c1c2cn]T is the vector of coefficients in the linear combination

(15)x(t)=c1x1(t)+c2x2(t)++cnxn(t).

Suppose now that we wish to solve the initial value problem

(21)dxdt=Px,x(a)=b,

where the initial vector b=[b1b2bn]T is given. Then, according to Eq. (19), it suffices to solve the system

(22)X(a)c=b

to find the coefficients c1, c2, , cn in Eq. (15). For instance, we can use the row-reduction techniques of Sections 3.2 and 3.3.

Example 4

Use the solution vectors given in Example 3 to solve the initial value problem

(23)dxdt=[320132013]x,x(0)=[026].

Solution

It follows from Theorem 3 that the linear combination

x(t)=c1x1(t)+c2x2(t)+c3x3(t)=c1[2et2etet]+c2[2e3t0e3t]+c3[2e5t2e5te5t]

is a general solution of the 3×3 linear system in (23). In scalar form, this gives the general solution

x1(t)=2c1et+2c2e3t+2c3e5t,x2(t)=2c1et2c3e5t,x3(t)=c1etc2e3t+c3e5t.

We seek the particular solution satisfying the initial conditions

x1(0)=0,x2(0)=2,x3(0)=6.

When we substitute these values in the preceding three scalar equations, we get the algebraic linear system

2c1+2c2+2c3=0,2c12c3=2,c1c2+c3=6

having the augmented coefficient matrix

[222020221116].

Multiplication of each of the first two rows by 12 gives

[111010111116];

then subtraction of the first row both from the second row and from the third row gives the matrix

[111001210206].

The first column of this matrix now has the desired form.

Now, we multiply the second row by 1, then add twice the result to the third row. Thereby, we get the upper triangular augmented coefficient matrix

[111001210044]

that corresponds to the transformed system

c1+c2+c3=0,c2+2c3=1,4c3=4.

We finally solve in turn for c3=1, c2=3, and c1=2. Thus the desired particular solution is given by

x(t)=2x1(t)3x2(t)+x3(t)=[4et6e3t+2e5t4et2e5t2et+3e3t+e5t].

Nonhomogeneous Solutions

We finally turn our attention to a nonhomogeneous linear system of the form

(24)dxdt=P(t)x+f(t).

The following theorem is analogous to Theorem 5 of Section 5.2 and is proved in precisely the same way, substituting the preceding theorems in this section for the analogous theorems of Section 5.2. In brief, Theorem 4 means that the general solution of Eq. (24) has the form

(25)x(t)=xc(t)+xp(t),

where xp(t) is a single particular solution of Eq. (24) and the complementary function xc(t) is a general solution of the associated homogeneous equation x=P(t)x.

Thus, finding a general solution of a homogeneous linear system involves two separate steps:

  1. Finding the general solution xc(t) of the associated homogeneous system;

  2. Finding a single particular solution xp(t) of the nonhomogeneous system.

The sum x(t)=xc(t)+xp(t) will then be a general solution of the nonhomogeneous system.

Example 5

The nonhomogeneous linear system

x1=3x12x29t+13,x2=x1+3x22x3+7t15,x3=x2+3x36t+7

can be recast as a matrix equation by writing

dxdt=[x1x2x3]=[3x12x29t+13x1+3x22x3+7t15x2+3x36t+7]=[3x12x2x1+3x22x3x2+3x2]+[9t+137t156t+7]=[320132013][x1x2x3]+[9t+137t156t+7].

Thus we get the form in (24) with

P(t)=[320132013],f(t)=[9t+137t156t+7].

In Example 3 we saw that a general solution of the associated homogeneous linear system

dxdt=[320132013]x

is given by

xc(t)=[2c1et+2c2e3t+2c3e5t2c1et2c3e5tc1etc2e3t+c3e5t],

and we can verify, by substitution, that the function

xp(t)=[3t52t]

(found by using a computer algebra system, or perhaps by a human being using a method discussed in Section 8.2) is a particular solution of the original nonhomogeneous system. Consequently, Theorem 4 implies that a general solution of the nonhomogeneous system is given by

x(t)=xc(t)+xp(t)

—that is, by

x1(t)=2c1et+2c2e3t+2c3e5t+3t,x2(t)=2c1et2c3e5t+5,x3(t)=c1etc2e3t+c3e5t+2t.

7.2 Problems

In Problems 1 and 2, verify the product law for differentiation, (AB)=AB+AB.

  1. A(t)=[t2t1t31t] and B(t)=[1t1+t3t24t3].

  2. A(t)=[ettt2t028t1t3] and B(t)=[32et3t].

In Problems 3 through 12, write the given system in the form x=P(t)x+f(t).

  1. x=3y, y=3x

  2. x=3x2y, y=2x+y

  3. x=2x+4y+3et, y=5xyt2

  4. x=txety+cos t, y=etx+t2ysin t

  5. x=y+z, y=z+x, z=x+y

  6. x=2x3y, y=x+y+2z, z=5y7z

  7. x=3x4y+z+t, y=x3z+t2, z=6y7z+t3

  8. x=txy+etz, y=2x+t2yz, z=etx+3ty+t3z

  9. x1=x2x2=2x3x3=3x4x4=4x1

  10. x1=x2+x3=1,x2=x3+x4=t,x3=x1+x4+t2,x4=x1+x2+t3

In Problems 13 through 22, first verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system.

  1. x=[4231]x; x1=[2et3et], x2=[e2te2t]

  2. x=[3234]x; x1=[e3t3e3t], x2=[2e2te2t]

  3. x=[3153]x; x1=e2t[11], x2=e2t[15]

  4. x=[4121]x; x1=e3t[11], x2=e2t[12]

  5. x=[4367]x; x1=[3e2t2e2t], x2=[e5t3e5t]

  6. x=[320132013]x; x1=et[221],x2=e3t[201], x3=e5t[221]

  7. x=[011101110]x; x1=e2t[111],x2=et[101], x3=et[011]

  8. x=[121610121]x; x1=[1613],x2=e3t[232], x3=e4t[121]

  9. x=[8112692661]x; x1=e2t[322],x2=et[111], x3=e3t[110]

  10. x=[14020100612160401]x; x1=et[1001],x2=et[0010], x3=et[0102], x4=et[1030]

In Problems 23 through 32, find a particular solution of the indicated linear system that satisfies the given initial conditions.

  1. The system of Problem 14: x1(0)=0, x2(0)=5

  2. The system of Problem 15: x1(0)=5, x2(0)=3

  3. The system of Problem 16: x1(0)=11, x2(0)=7

  4. The system of Problem 17: x1(0)=8, x2(0)=0

  5. The system of Problem 18: x1(0)=0, x2(0)=0, x3(0)=4

  6. The system of Problem 19: x1(0)=10, x2(0)=12, x3(0)=1

  7. The system of Problem 21: x1(0)=1, x2(0)=2, x3(0)=3

  8. The system of Problem 21: x1(0)=5, x2(0)=7, x3(0)=11

  9. The system of Problem 22: x1(0)=x2(0)=x3(0)=x4(0)=1

  10. The system of Problem 22: x1(0)=1, x2(0)=3, x3(0)=4, x4(0)=7

    1. Show that the vector functions

      x1(t)=[tt2]andx2=[t2t3]

      are linearly independent on the real line.

    2. Why does it follow from Theorem 2 that there is no continuous matrix P(t) such that x1 and x2 are both solutions of x=P(t)x?

  11. Suppose that one of the vector functions

    x1(t)=[x11(t)x21(t)]andx2(t)=[x12(t)x22(t)]

    is a constant multiple of the other on the open interval I. Show that their Wronskian W(t)=|[xij(t)]| must vanish identically on I. This proves part (a) of Theorem 2 in the case n=2.

  12. Suppose that the vectors x1(t) and x2(t) of Problem 34 are solutions of the equation x=P(t)x, where the 2×2 matrix P(t) is continuous on the open interval I. Show that if there exists a point a of I at which their Wronskian W(a) is zero, then there exist numbers c1 and c2 not both zero such that c1x1(a)+c2x2(a)=0. Then conclude from the uniqueness of solutions of the equation x=P(t)x that

    c1x1(t)+c2x2(t)=0

    for all t in I; that is, that x1 and x2 are linearly dependent. This proves part (b) of Theorem 2 in the case n=2.

  13. Generalize Problems 34 and 35 to prove Theorem 2 for n an arbitrary positive integer.

  14. Let x1(t), x2(t), , xn(t) be vector functions whose ith components (for some fixed i) xi1(t), xi2(t), , xin(t) are linearly independent real-valued functions. Conclude that the vector functions are themselves linearly independent.

7.3 The Eigenvalue Method for Linear Systems

We now introduce a powerful method for constructing the general solution of a homogeneous first-order system with constant coefficients,

(1)x1=a11x1+a12x2++a1nxn,x2=a21x1+a22x2++a2nxn,xn=an1x1+an2x2++annxn.

By Theorem 3 of Section 7.2, we know that it suffices to find n linearly independent solution vectors x1, x2, , xn; the linear combination

(2)x(t)=c1x1+c2x2++cnxn

with arbitrary coefficients will then be a general solution of the system in (1).

To search for the n needed linearly independent solution vectors, we proceed by analogy with the characteristic root method for solving a single homogeneous equation with constant coefficients (Section 5.3). It is reasonable to anticipate solution vectors of the form

(3)x(t)=[x1x2x3xn]=[v1eλtv2eλtv3eλtvneλt]=[v1v2v3vn]eλt=veλt,

where λ, v1, v2, v3, , vn are appropriate scalar constants. For if we substitute

xi=vieλt,xi=λvieλt(i=1,2,,n)

in (1), then each term in the resulting equations will have the factor eλt, so we can cancel it throughout. This will leave us with n linear equations that—for appropriate values of λ—we can hope to solve for values of the coefficients v1, v2, , vn in Eq. (3), so that x(t)=veλt is, indeed, a solution of the system in (1).

To investigate this possibility, it is more efficient to write the system in (1) in the matrix form

(4)x=Ax,

where A=[aij]. When we substitute the trial solution x=veλt (having derivative x=λveλt) in Eq. (4), the result is

λveλt=Aveλt.

We cancel the nonzero scalar factor eλt to get

(5)Av=λv.

Recall, from Section 6.1, that Eq. (5) means that v0 is an eigenvector of the matrix A associated with the eigenvalue λ. Our discussion of Eqs. (3)(5) therefore provides a proof of the following theorem, which is the basis for the eigenvalue method of solving a first-order system with constant coefficients.

Recall that an eigenvalue λ of the matrix A is a solution of the characteristic equation

(6)|AλI|=det(AλI)=0

and that an eigenvector v associated with λ is then a solution of the eigenvector equation

(7)(AλI)v=0.

The Eigenvalue Method

In outline, this method for solving the n×n homogeneous constant-coefficient system x=Ax proceeds as follows:

  1. First, we solve the characteristic equation in (6) for the eigenvalues λ1, λ2, , λn of the matrix A.

  2. Next, we attempt to find n linearly independent eigenvectors v1, v2, , vn associated with these eigenvalues.

  3. Step 2 is not always possible, but, when it is, we get n linearly independent solutions

    (8)x1(t)=v1eλit,x2(t)=v2eλ2t,,xn(t)=vneλnt.

    In this case the general solution of x=Ax is a linear combination

    x(t)=c1x1(t)+c2x2(t)++cnxn(t)

    of these n solutions.

We will discuss separately the various cases that can occur, depending on whether the eigenvalues are distinct or repeated, real or complex. The case of repeated eigenvalues—multiple roots of the characteristic equation—will be deferred to Section 7.6.

Distinct Real Eigenvalues

If the eigenvalues λ1, λ2, , λn are real and distinct, then we substitute each of them in turn in Eq. (7) and solve for the associated eigenvectors v1, v2, , vn. It then follows from Theorem 2 in Section 6.2 that the particular solutions given in (8) are linearly independent. (In any particular example, such linear independence can always be verified by using the Wronskian determinant of Section 7.2.) The following example illustrates the eigenvalue method.

Example 1

Find a general solution of the system

(9)x1=4x1+2x2,x2=3x1x2.

Solution

The matrix form of the system in (9) is

(10)x=[4231]x.

The characteristic equation of the coefficient matrix is

|4λ231λ|=(4λ)(1λ)6=λ23λ10=(λ+2)(λ5)=0,

so we have the distinct real eigenvalues λ1=2 and λ2=5.

For the coefficient matrix A in Eq. (10), the eigenvector equation (AλI)v=0 takes the form

(11)[4λ231λ][ab]=[00]

for the associated eigenvector v=[ab]T.

Case 1: λ1=2. Substitution of the first eigenvalue λ1=2 in Eq. (11) yields the system

[6231][ab]=[00]

—that is, the two scalar equations

(12)6a+2b=0,3a+b=0.

In contrast with the typical nonsingular (algebraic) linear system that has a unique solution, the homogeneous linear system in (12) is singular—the two scalar equations obviously are equivalent (each being a multiple of the other). Therefore, Eq. (12) has infinitely many nonzero solutions—we can choose a arbitrarily (but nonzero) and then solve for b.

Substitution of an eigenvalue λ in the eigenvector equation (AλI)v=0 always yields a singular homogeneous linear system, and among its infinity of solutions we generally seek a “simple” solution with small integer values (if possible). Looking at the second equation in (12), the choice a=1 yields b=3, and thus

v1=[13]

is an eigenvector associated with λ1=2 (as is any nonzero constant multiple of v1).

Remark

If, instead of the “simplest” choice a=1, b=3, we had made another choice a=c, b=3c, we would have obtained the eigenvector

v1=[c3c]=c[13].

Because this is a constant multiple of our previous result, any choice we make leads to (a constant multiple of ) the same solution

x1(t)=[13]e2t.

Case 2: λ2=5. Substitution of the second eigenvalue λ=5 in (11) yields the pair

(13)a+2b=0,3a6b=0

of equivalent scalar equations. With b=1 in the first equation we get a=2, so

v2=[21]

is an eigenvector associated with λ2=5. A different choice—a=2c, b=c—would merely give a [constant] multiple of v2.

These two eigenvalues and their associated eigenvectors yield the two solutions

x1(t)=[13]e2tandx2(t)=[21]e5t.

They are linearly independent, because their Wronskian

|e2t2e5t3e2te5t|=7e3t

is nonzero. Hence a general solution of the system in (10) is

x(t)=c1x1(t)+c2x2(t)=c1[13]e2t+c2[21]e5t;

in scalar form,

x1(t)=c1e2t+2c2e5t,x2(t)=3c1e2t+c2e5t.

Figure 7.3.1 shows some typical solution curves of the system in (10). We see two families of hyperbolas sharing the same pair of asymptotes: the line x1=2x2 obtained from the general solution with c1=0, and the line x2=3x1 obtained with c2=0. Given initial values x1(0)=b1, x2(0)=b2, it is apparent from the figure that

  • If (b1,b2) lies to the right of the line x2=3x1, then x1(t) and x2(t) both tend to + as t+;

  • If (b1,b2) lies to the left of the line x2=3x1, then x1(t) and x2(t) both tend to as t+.

FIGURE 7.3.1.

Direction field and solution curves for the linear system x1=4x1+2x2, x2=3x1x2 of Example 1.

Remark

As in Example 1, it is convenient when discussing a linear system x=Ax to use vectors x1, x2, , xn to denote different vector-valued solutions of the system, whereas the scalars x1, x2, , xn denote the components of a single vector-valued solution x.

Compartmental Analysis

Frequently, a complex process or system can be broken down into simpler subsystems or “compartments” that can be analyzed separately. The whole system can then be modeled by describing the interactions between the various compartments. Thus a chemical plant may consist of a succession of separate stages (or even physical compartments) in which various reactants and products combine or are mixed. It may happen that a single differential equation describes each compartment of the system; then the whole physical system is modeled by a system of differential equations.

As a simple example of a three-stage system, Fig. 7.3.2 shows three brine tanks containing V1, V2, and V3 gallons of brine, respectively. Fresh water flows into tank 1, while mixed brine flows from tank 1 into tank 2, from tank 2 into tank 3, and out of tank 3. Let xi(t) denote the amount (in pounds) of salt in tank i at time t, for i=1, 2, and 3. If each flow rate is r gallons per minute, then a simple accounting of salt concentrations, as in Example 2 of Section 7.1, yields the first-order system

(14)x1=k1x1,x2=k1x1k2x2,x3=k2x2k3x3,

where

(15)ki=rVi,i=1,2,3.

Example 2

Three brine tanks If V1=20, V2=40, V3=50, r=10 (gal/min), and the initial amounts of salt in the three brine tanks, in pounds, are

x1(0)=15,x2(0)=x3(0)=0,

find the amount of salt in each tank at time t0.

Solution

Substituting the given numerical values in (14) and (15), we get the initial value problem

(16)x(t)=[0.50.00.00.50.250.00.00.250.2]x,x(0)=[1500]

for the vector x(t)=[x1(t)x2(t)x3(t)]T. The simple form of the matrix

(17)AλI=[0.5λ0.00.00.50.25λ0.00.00.250.2λ]

leads readily to the characteristic equation

|AλI|=(0.5λ)(0.25λ)(0.2λ)=0.

Thus, the coefficient matrix A in (16) has the distinct eigenvalues λ1=0.5, λ2=0.25, and λ3=0.2 and therefore has three linearly independent eigenvectors.

FIGURE 7.3.2.

The three brine tanks of Example 2.

Case 1: λ1=0.5. Substituting λ=0.5 in (17), we get the equation

[A+(0.5)I]v=[0.00.00.00.50.250.00.00.250.3][abc]=[000]

for the associated eigenvector v=[abc]T. The last two rows, after division by 0.25 and 0.05, respectively, yield the scalar equations

2a+b=0,5b+6c=0.

The second equation is satisfied by b=6 and c=5, and then the first equation gives a=3. Thus the eigenvector

v1=[365]T

is associated with the eigenvalue λ1=0.5.

Case 2: λ2=0.25. Substituting λ=0.25 in (17), we get the equation

[A+(0.25)I]v=[0.25000.50000.250.05][abc]=[000]

for the associated eigenvector v=[abc]T. Each of the first two rows implies that a=0, and division of the third row by 0.05 gives the equation

5b+c=0,

which is satisfied by b=1, c=5. Thus the eigenvector

v2=[015]T

is associated with the eigenvalue λ2=0.25.

Case 3: λ3=0.2. Substituting λ=0.2 in (17), we get the equation

[A+(0.2)I]v=[0.30.00.00.50.050.00.00.250.0][abc]=[000]

for the eigenvector v. The first and third rows imply that a=0, and b=0, respectively, but the all-zero third column leaves c arbitrary (but nonzero). Thus

v3=[001]T

is an eigenvector associated with λ3=0.2.

The general solution

x(t)=c1v1eλ1t+c2v2eλ2t+c3v3eλ3t

therefore takes the form

x(t)=c1[363]e(0.5)t+c2[015]e(0.25)t+c3[001]e(0.2)t.

The resulting scalar equations are

x1(t)=3c1e(0.5)t,x2(t)=6c1e(0.5)t+c2e(0.25)t,x3(t)=5c1e(0.5)t5c2e(0.25)t+c3e(0.2)t.

When we impose the initial conditions x1(0)=15, x2(0)=x3(0)=0, we get the equations

3c1=15,6c1+c2=0,5c15c2+c3=0

that are readily solved (in turn) for c1=5, c2=30, and c3=125. Thus, finally, the amounts of salt at time t in the three brine tanks are given by

x1(t)=15e(0.5)t,x2(t)=30e(0.5)t+30e(0.25)t,x3(t)=25e(0.5)t150e(0.25)t+125e(0.2)t.

Figure 7.3.3 shows the graphs of x1(t), x2(t), and x3(t). As we would expect, tank 1 is rapidly “flushed” by the incoming fresh water, and x1(t)0 as t+. The amounts x2(t) and x3(t) of salt in tanks 2 and 3 peak in turn and then approach zero as the whole three-tank system is purged of salt as t+.

FIGURE 7.3.3.

The salt content functions of Example 2.

Complex Eigenvalues

Even if some of the eigenvalues are complex, so long as they are distinct the method described previously still yields n linearly independent solutions. The only complication is that the eigenvectors associated with complex eigenvalues are ordinarily complex-valued, so we will have complex-valued solutions.

To obtain real-valued solutions, we note that—because we are assuming that the matrix A has only real entries—the coefficients in the characteristic equation will all be real. Consequently any complex eigenvalues must appear in complex-conjugate pairs. Suppose, then, that λ=p+qi and λ̲=pqi are such a pair of eigenvalues. If v is an eigenvector associated with λ, so that

(AλI)v=0,

then taking complex conjugates in this equation yields

(Aλ¯I)v¯=0

since A̲=A and I̲=I (these matrices being real) and the conjugate of a complex product is the product of the conjugates of the factors. Thus the conjugate v̲ of v is an eigenvector associated with λ̲. Of course, the conjugate of a vector is defined componentwise; if

(18)v=[a1+b1ia2+b2ian+bni]=[a1a2an]+[b1b2bn]i=a+bi,

then v̲=abi. The complex-valued solution associated with λ and v is then

x(t)=veλt=ve(p+qi)t=(a+bi)ept(cosqt+isinqt)

—that is,

(19)x(t)=ept(acosqtbsinqt)+iept(bcosqt+asinqt).

Because the real and imaginary parts of a complex-valued solution are also solutions, we thus get the two real-valued solutions

(20)x1(t)=Re[x(t)]=ept(acosqtbsinqt),x2(t)=Im[x(t)]=ept(bcosqt+asinqt)

associated with the complex conjugate eigenvalues p±qi. It is easy to check that the same two real-valued solutions result from taking real and imaginary parts of v̲eλ̲t. Rather than memorizing the formulas in (20), it is preferable in a specific example to proceed as follows:

Example 3

Find a general solution of the system

(21)dx1dt=4x13x2,dx2dt=3x1+4x2.

Solution

The coefficient matrix

A=[4334]

has the characteristic equation

|AλI|=|4λ334λ|=(4λ)2+9=0

and hence has the complex conjugate eigenvalues λ=43i and λ̲=4+3i.

Substituting λ=43i in the eigenvector equation (AλI)v=0, we get the equation

[A(43i)I]v=[3i333i][ab]=[00]

for an associated eigenvalue v=[ab]T. Division of each row by 3 yields the two scalar equations

iab=0,a+ib=0,

each of which is satisfied by a=1 and b=i. Thus, v=[1i]T is a complex eigenvector associated with the complex eigenvalue λ=43i.

The corresponding complex-valued solution x(t)=veλt of x=Ax is then

x(t)=[1i]e(43i)t=[1i]e4t(cos3tisin3t)=e4t[cos3tisin3ticos3t+sin3t].

The real and imaginary parts of x(t) are the real-valued solutions

x1(t)=e4t[cos3tsin3t]andx2(t)=e4t[sin3tcos3t].

A real-valued general solution of x=Ax is then given by

x(t)=c1x1(t)+c2x2(t)=e4t[c1cos3tc2sin3tc1sin3t+c2cos3t].

Finally, a general solution of the system in (21) in scalar form is

x1(t)=e4t(c1cos3tc2sin3t),x2(t)=e4t(c1sin3t+c2cos3t).

Figure 7.3.4 shows some typical solution curves of the system in (21). Each appears to spiral counterclockwise as it emanates from the origin in the x1x2-plane. Actually, because of the factor e4t in the general solution, we see that

  • Along each solution curve, the point (x1(t),x2(t)) approaches the origin as t, whereas

  • The absolute values of x1(t) and x2(t) both increase without bound as t+.

FIGURE 7.3.4.

Direction field and solution curves for the linear system x1=4x13x2, x2=3x1+4x2 of Example 3.

Figure 7.3.5 shows a “closed” system of three brine tanks with volumes V1, V2, and V3. The difference between this system and the “open” system of Fig. 7.3.2 is that now the inflow to tank 1 is the outflow from tank 3. With the same notation as in Example 2, the appropriate modification of Eq. (14) is

(22)dx1dt=k1x1+k3x3,dx2dt=k1x1k2x2,dx3dt=k2x2k3x3,

where ki=r/Vi as in (15).

FIGURE 7.3.5.

The three brine tanks of Example 4.

Example 4

Find the amounts x1(t), x2(t), and x3(t) of salt at time t in the three brine tanks of Fig. 7.3.5 if V1=50 gal, V2=25 gal, V3=50 gal, and r=10 gal/min.

Solution

With the given numerical values, (22) takes the form

(23)dxdt=[0.200.20.20.4000.40.2]x,

where x=[x1x2x3]T as usual. When we expand the determinant of the matrix

(24)AλI=[0.2λ0.00.20.20.4λ0.00.00.40.2λ]

along its first row, we find that the characteristic equation of A is

(0.2λ)(0.4λ)(0.2λ)+(0.2)(0.2)(0.4)=λ3(0.8)λ2(0.2)λ=λ[(λ+0.4)2+(0.2)2]=0.

Thus A has the zero eigenvalue λ0=0 and the complex conjugate eigenvalues λ, λ̲=0.4±(0.2)i. We anticipate one solution corresponding to the zero eigenvalue and two additional linearly independent solutions corresponding to the complex conjugate eigenvalues.

Case 1: λ0=0. Substitution of λ=0 in Eq. (24) gives the eigenvector equation

(25)(A0I)v=[0.20.00.20.20.40.00.00.40.2][abc]=[000]

for v=[abc]T. The first row gives a=c and the second row gives a=2b, so v0=[212]T is an eigenvector associated with the eigenvalue λ0=0. The corresponding solution x0(t)=v0eλ0t of Eq. (23) is the constant solution

(26)x0(t)=[212].

Case 2: λ=0.4(0.2)i. Substitution of λ=0.4(0.2)i in Eq. (24) gives the eigenvector equation

[A(0.4(0.2)i)I]v=[0.2+(0.2)i0.00.20.2(0.2)i0.00.00.40.2+(0.2)i][abc]=[000].

The second equation (0.2)a+(0.2)ib=0 is satisfied by a=1 and b=i. Then the first equation

[0.2+(0.2)i]a+(0.2)c=0

gives c=1i. Thus, v=[1i(1i)]T is a complex eigenvector associated with the complex eigenvalue λ=0.4(0.2)i.

The corresponding complex-valued solution x(t)=veλt of (23) is

x(t)=[1i1i]Te(0.40.2i)t=[1i1i]Te(0.4)t(cos 0.2ti sin 0.2t)=e(0.4)t[cos 0.2ti sin 0.2tsin 0.2t+i cos 0.2tcos 0.2tsin 0.2ti cos 0.2t+i sin 0.2t].

The real and imaginary parts of x(t) are the real-valued solutions

(27)x1(t)=e(0.4)t[cos 0.2tsin 0.2tcos 0.2tsin 0.2t],x2(t)=e(0.4)t[sin 0.2tcos 0.2tcos 0.2t+sin 0.2t].

The general solution

x(t)=c0x0(t)+c1x1(t)+c2x2(t)

has scalar components

(28)x1(t)=2c0+e(0.4)t(c1 cos 0.2tc2 sin 0.2t),x2(t)=c0+e(0.4)t(c1 sin 0.2t+c2 cos 0.2t),x3(t)=2c0+e(0.4)t[(c1c2) cos 0.2t+(c1+c2) sin 0.2t]

giving the amounts of salt in the three tanks at time t.

Observe that

(29)x1(t)+x2(t)+x3(t)5c0.

Of course, the total amount of salt in the closed system is constant; the constant c0 in (29) is one-fifth the total amount of salt. Because of the factors of e(0.4)t in (28), we see that

limtx1(t)=2c0,limtx2(t)=c0,andlimtx3(t)=2c0.

Thus, as t+ the salt in the system approaches a steady-state distribution with 40% of the salt in each of the two 50-gallon tanks and 20% in the 25-gallon tank. So whatever the initial distribution of salt among the three tanks, the limiting distribution is one of uniform concentration throughout the system. Figure 7.3.6 shows the graphs of the three solution functions with c0=10,c1=30, and c2=10, in which case

x1(0)=50andx2(0)=x3(0)=0.

FIGURE 7.3.6.

The salt content functions of Example 4.

7.3 Problems

In Problems 1 through 16, apply the eigenvalue method of this section to find a general solution of the given system. If initial values are given, find also the corresponding particular solution. For each problem, use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system.

  1. x1=x1+2x2,x2=2x1+x2

  2. x1=2x1+3x2,x2=2x1+x2

  3. x1=3x1+4x2,x2=3x1+2x2;x1(0)=x2(0)=1

  4. x1=4x1+x2,x2=6x1x2

  5. x1=6x17x2,x2=x12x2

  6. x1=9x1+5x2,x2=6x12x2;x1(0)=1,x2(0)=0

  7. x1=3x1+4x2,x2=6x15x2

  8. x1=x15x2,x2=x1x2

  9. x1=2x15x2,x2=4x12x2;x1(0)=2,x2(0)=3

  10. x1=3x12x2,x2=9x13x2

  11. x1=x12x2,x2=2x1+x2;x1(0)=0,x2(0)=4

  12. x1=x15x2,x2=x1+3x2

  13. x1=5x19x2,x2=2x1x2

  14. x1=3x14x2,x2=4x1+3x2

  15. x1=7x15x2,x2=4x1+3x2

  16. x1=50x1+20x2,x2=100x160x2

In Problems 17 through 25, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of each system.

  1. x1=4x1+x2+4x3, x2=x1+7x2+x3, x3=4x1+x2+4x3

  2. x1=x1+2x2+2x3, x2=2x1+7x2+x3, x3=2x1+x2+7x3

  3. x1=4x1+x2+x3,x2=x1+4x2+x3,x3=x1+x2+4x3

  4. x1=5x1+x2+3x3, x2=x1+7x2+x3, x3=3x1+x2+5x3

  5. x1=5x16x3, x2=2x1x22x3, x3=4x12x24x3

  6. x1=3x1+2x2+2x3, x2=5x14x22x3, x3=5x1+5x2+3x3

  7. x1=3x1+x2+x3, x2=5x13x2x3, x3=5x1+5x2+3x3

  8. x1=2x1+x2x3, x2=4x13x2x3, x3=4x1+4x2+2x3

  9. x1=5x1+5x2+2x3, x2=6x16x25x3, x3=6x1+6x2+5x3

  10. Find the particular solution of the system

    dx1dt=3x1+x3,dx2dt=9x1x2+2x3,dx3dt=9x1+4x2x3

    that satisfies the initial conditions x1(0)=0, x2(0)=0, x3(0)=17.

Cascading Brine Tanks

The amounts x1(t) and x2(t) of salt in the two brine tanks of Fig. 7.3.7 satisfy the differential equations

dx1dt=k1x1,dx2dt=k1x1k2x2,

where ki=r/Vi for i=1, 2. In Problems 27 and 28 the volumes V1 and V2 are given. First solve for x1(t) and x2(t), assuming that r=10 (gal/min), x1(0)=15 (lb), and x2(0)=0. Then find the maximum amount of salt ever in tank 2. Finally, construct a figure showing the graphs of x1(t) and x2(t).

FIGURE 7.3.7.

The two brine tanks of Problems 27 and 28.

  1. V1=50 (gal), V2=25 (gal)

  2. V1=25 (gal), V2=40 (gal)

Interconnected Brine Tanks

The amounts x1(t) and x2(t) of salt in the two brine tanks of Fig. 7.3.8 satisfy the differential equations

dx1dt=k1x1+k2x2,dx2dt=k1x1k2x2,

where ki=r/Vi as usual. In Problems 29 and 30, solve for x1(t) and x2(t), assuming that r=10 (gal/min), x1(0)=15 (lb), and x2(0)=0. Then construct a figure showing the graphs of x1(t) and x2(t).

FIGURE 7.3.8.

The two brine tanks of Problems 29 and 30.

  1. V1=50 (gal), V2=25 (gal)

  2. V1=25 (gal), V2=40 (gal)

Open Three-Tank System

Problems 31 through 34 deal with the open three-tank system of Fig. 7.3.2. Fresh water flows into tank 1; mixed brine flows from tank 1 into tank 2, from tank 2 into tank 3, and out of tank 3; all at the given flow rate r gallons per minute. The initial amounts x1(0)=x0 (lb), x2(0)=0, and x3(0)=0 of salt in the three tanks are given, as are their volumes V1, V2, and V3 (in gallons). First solve for the amounts of salt in the three tanks at time t, then determine the maximal amount of salt that tank 3 ever contains. Finally, construct a figure showing the graphs of x1(t), x2(t), and x3(t).

  1. r=30, x0=27, V1=30, V2=15, V3=10

  2. r=60, x0=45, V1=20, V2=30, V3=60

  3. r=60, x0=45, V1=15, V2=10, V3=30

  4. r=60, x0=40, V1=20, V2=12, V3=60

Closed Three-Tank System

Problems 35 through 37 deal with the closed three-tank system of Fig. 7.3.5, which is described by the equations in (24). Mixed brine flows from tank 1 into tank 2, from tank 2 into tank 3, and from tank 3 into tank 1, all at the given flow rate r gallons per minute. The initial amounts x1(0)=x0 (pounds), x2(0)=0, and x3(0)=0 of salt in the three tanks are given, as are their volumes V1, V2, and V3 (in gallons). First solve for the amounts of salt in the three tanks at time t, then determine the limiting amount (as t+) of salt in each tank. Finally, construct a figure showing the graphs of x1(t), x2(t), and x3(t).

  1. r=120, x0=33, V1=20, V2=6, V3=40

  2. r=10, x0=18, V1=20, V2=50, V3=20

  3. r=60, x0=55, V1=60, V2=20, V3=30

For each matrix A given in Problems 38 through 40, the zeros in the matrix make its characteristic polynomial easy to calculate. Find the general solution of x=Ax.

  1. A=[1000220003300044]

  2. A=[20094201000180001]

  3. A=[2000215279005000212]

  4. The coefficient matrix A of the 4×4 system

    x1=4x1+x2+x3+7x4,x2=x1+4x2+10x3+x4,x3=x1+10x2+4x3+x4,x4=7x1+x2+x3+4x4

    has eigenvalues λ1=3, λ2=6, λ3=10, and λ4=15. Find the particular solution of this system that satisfies the initial conditions

    x1(0)=3,x2(0)=x3(0)=1,x4(0)=3.

In Problems 42 through 50, use a calculator or computer system to calculate the eigenvalues and eigenvectors (as illustrated in Application 7.3) in order to find a general solution of the linear system x=Ax with the given coefficient matrix A.

  1. A=[40125435134625734]

  2. A=[2011131217482131]

  3. A=[147232029091299015123]

  4. A=[975012711924171991813179]

  5. A=[13421061392165270162031162233]

  6. A=[23181608679342726926212512]

  7. A=[4785510321821394016712123264360248]

  8. A=[1391452142822578737038139387615216591335951038723]

  9. A=[9130001314191020104301273012181210109102690651514231020100]

7.3 Application Automatic Calculation of Eigenvalues and Eigenvectors

Most computational systems offer the capability to find eigenvalues and eigenvectors readily. For instance, Fig. 7.3.9 shows a graphing calculator computation of the eigenvalues and eigenvectors of the matrix

A=[0.50.00.00.50.250.00.00.250.2]

FIGURE 7.3.9.

TI-Nspire CX CAS calculation of the eigenvalues and eigenvectors of the matrix A.

of Example 2. We see the three eigenvectors displayed as column vectors, appearing in the same order as their corresponding eigenvalues. In this display the eigenvectors are normalized, that is, multiplied by an appropriate scalar so as to have length 1. You can verify, for example, that the displayed eigenvector corresponding to the third eigenvalue λ=12 is a scalar multiple of v=[1253]T.. The Maple commands

with(linalg)
A := matrix(3,3,[−0.5,0,0,0.5,−0.25,0,0,0.25,−0.2]);
eigenvects(A);

the Mathematica commands

A = {{−0.5,0,0},{0.5,−0.25,0},{0,0.25,−0.2}}
Eigensystem[A]

the Wolfram|Alpha query

((−0.5, 0, 0), (0.5, −0.25, 0), (0, 0.25, −0.2))

and the Matlab commands

A = [−0.5,0,0; 0.5,−0.25,0; 0,0.25,−0.2]
[V,D] = eig(A)

(where D will be a diagonal matrix displaying the eigenvalues of A and the column vectors of V are the corresponding eigenvectors) produce similar results. You can use these commands to find the eigenvalues and eigenvectors needed for any of the problems in this section.

For a more substantial investigation, choose a positive integer n<10 (n=5, for instance) and let q1, q2, , qn denote the first n nonzero digits in your student ID number. Now consider an open system of brine tanks as in Fig. 7.3.2, except with n rather than three successive tanks having volumes Vi=10qi (i=1, 2, , n) in gallons. If each flow rate is r=10 gallons per minute, then the salt amounts x1(t), x2(t), , xn(t) satisfy the linear system

x1=k1x1,xi=ki1x11kix1(i=2,3,,n),

where ki=r/Vi. Apply the eigenvalue method to solve this system with initial conditions

x1(0)=10,x2(0)=x3(0)==xn(0)=0.

Graph the solution functions and estimate graphically the maximum amount of salt that each tank ever contains.

For an alternative investigation, suppose that the system of n tanks is closed as in Fig. 7.3.5, so that tank 1 receives as inflow the outflow from tank n (rather than fresh water). Then the first equation should be replaced with x1=knxnk1x1. Now show that, in this closed system, as t+ the salt originally in tank 1 distributes itself with constant density throughout the various tanks. A plot like Fig. 7.3.6 should make this fairly obvious.

7.4 A Gallery of Solution Curves of Linear Systems

In the preceding section we saw that the eigenvalues and eigenvectors of the n×n matrix A are of central importance to the solutions of the homogeneous linear constant-coefficient system

(1)x=Ax.

Indeed, according to Theorem 1 from Section 7.3, if λ is an eigenvalue of A and v is an eigenvector of A associated with λ, then

(2)x(t)=veλt

is a nontrivial solution of the system (1). Moreover, if A has n linearly independent eigenvectors v1, v2, , vn associated with its n eigenvalues λ1, λ2, , λn, then in fact all solutions of the system (1) are given by linear combinations

(3)x(t)=c1v1eλ1t+c2v2eλ2t++cnvneλnt,

where c1,c2,,cn are arbitrary constants. If the eigenvalues include complex conjugate pairs, then we can obtain a real-valued general solution from Eq. (3) by taking real and imaginary parts of the terms in (3) corresponding to the complex eigenvalues.

Our goal in this section is to gain a geometric understanding of the role that the eigenvalues and eigenvectors of the matrix A play in the solutions of the system (1). We will see, illustrating primarily with the case n=2, that particular arrangements of eigenvalues and eigenvectors correspond to identifiable patterns—“fingerprints,” so to speak—in the phase plane portrait of the system (1). Just as in algebra we learn to recognize when an equation in x and y corresponds to a line or parabola, we can predict the general appearance of the solution curves of the system (1) from the eigenvalues and eigenvectors of the matrix A. By considering various cases for these eigenvalues and eigenvectors we will create a “gallery”—Figure 7.4.16 appearing at the end of this section—of typical phase plane portraits that gives, in essence, a complete catalog of the geometric behaviors that the solutions of a 2×2 homogeneous linear constant-coefficient system can exhibit. This will help us analyze not only systems of the form (1), but also more complicated systems that can be approximated by linear systems, a topic we explore in Section 9.2.

Systems of Dimension n=2

Until stated otherwise, we henceforth assume that n=2, so that the eigenvalues of the matrix A are λ1 and λ2. According to Theorem 2 of Section 6.2, if λ1 and λ2 are distinct, then the associated eigenvectors v1 and v2 of A are linearly independent. In this event, the general solution of the system (1) is given by

(4)x(t)=c1v1eλ1t+c2v2eλ2t

if λ1 and λ2 are real, and by

(5)x(t)=c1ept(acosqtbsinqt)+c2ept(bcosqt+asinqt)

if λ1 and λ2 are the complex conjugate numbers p±iq; here the vectors a and b are the real and imaginary parts, respectively, of a (complex-valued) eigenvector of A associated with the eigenvalue p±iq. If instead λ1 and λ2 are equal (to a common value λ, say), then as we will see in Section 7.6, the matrix A may or may not have two linearly independent eigenvectors v1 and v2. If it does, then the eigenvalue method of Section 7.3 applies once again, and the general solution of the system (1) is given by the linear combination

(6)x(t)=c1v1eλt+c2v2eλt

as before. If A does not have two linearly independent eigenvectors, then—as we will see—we can find a vector v2 such that the general solution of the system (1) is given by

(7)x(t)=c1v1eλt+c2(v1t+v2)eλt,

where v1 is an eigenvector of A associated with the lone eigenvalue λ. The nature of the vector v2 and other details of the general solution in (7) will be discussed in Section 7.6, but we include this case here in order to make our gallery complete.

With this algebraic background in place, we begin our analysis of the solution curves of the system (1). First we assume that the eigenvalues λ1 and λ2 of the matrix A are real, and subsequently we take up the case where λ1 and λ2 are complex conjugates.

Real Eigenvalues

We will divide the case where λ1 and λ2 are real into the following possibilities:

Distinct eigenvalues

Repeated eigenvalue

Saddle Points

Nonzero Distinct Eigenvalues of Opposite Sign: The key observation when λ1<0<λ2 is that the positive scalar factors eλ1t and eλ2t in the general solution

(4)x(t)=c1v1eλ1t+c2v2eλ2t

of the system x=Ax move in opposite directions (on the real line) as t varies. For example, as t grows large and positive, eλ2t grows large, because λ2>0, whereas eλ1t approaches zero, because λ1<0; thus the term c1v1eλ1t in the solution x(t) in (4) vanishes and x(t) approaches c2v2eλ2t. If instead t grows large and negative, then the opposite occurs: The factor eλ1t grows large whereas eλ2t becomes small, and the solution x(t) approaches c1v1eλ1t. If we assume for the moment that both c1 and c2 are nonzero, then loosely speaking, as t ranges from to +, the solution x(t) shifts from being “mostly” a multiple of the eigenvector v1 to being “mostly” a multiple of v2.

Geometrically, this means that all solution curves given by (4) with both c1 and c2 nonzero have two asymptotes, namely the lines l1 and l2 passing through the origin and parallel to the eigenvectors v1 and v2, respectively; the solution curves approach l1 as t and l2 as t+. Indeed, as Fig. 7.4.1 illustrates, the lines l1 and l2 effectively divide the plane into four “quadrants” within which all solution curves flow from the asymptote l1 to the asymptote l2 as t increases. (The eigenvectors shown in Fig. 7.4.1—and in other figures—are scaled so as to have equal length.) The particular quadrant in which a solution curve lies is determined by the signs of the coefficients c1 and c2. If c1 and c2 are both positive, for example, then the corresponding solution curve extends asymptotically in the direction of the eigenvector v1 as t, and asymptotically in the direction of v2 as t. If instead c1>0 but c2<0, then the corresponding solution curve still extends asymptotically in the direction of v1 as t, but extends asymptotically in the direction opposite v2 as t+ (because the negative coefficient c2 causes the vector c2v2 to point “backwards” from v2).

FIGURE 7.4.1.

Solution curves x(t)=c1v1eλ1t+c2v2eλ2t for the system x=Ax when the eigenvalues λ1,λ2 of A are real with λ1<0<λ2.

If c1 or c2 equals zero, then the solution curve remains confined to one of the lines l1 and l2. For example, if c10 but c2=0, then the solution (4) becomes x(t)=c1v1eλ1t, which means that the corresponding solution curve lies along the line l1. It approaches the origin as t+, because λ1<0, and recedes farther and farther from the origin as t, either in the direction of v1 (if c1>0) or the direction opposite v1 (if c1<0). Similarly, if c1=0 and c20, then because λ2>0, the solution curve flows along the line l2 away from the origin as t+ and toward the origin as t.

Figure 7.4.1 illustrates typical solution curves corresponding to nonzero values of the coefficients c1 and c2. Because the overall picture of the solution curves is suggestive of the level curves of a saddle-shaped surface (like z=xy), we call the origin a saddle point for the system x=Ax.

Example 1

The solution curves in Fig. 7.4.1 correspond to the choice

(8)A=[4161]

in the system x=Ax; as you can verify, the eigenvalues of A are λ1=2 and λ2=5 (thus λ1<0<λ2), with associated eigenvectors

v1=[16]andv2=[11].

According to Eq. (4), the resulting general solution is

(9)x(t)=c1[16]e2t+c2[11]e5t,

or, in scalar form,

(10)x1(t)=c1e2t+c2e5t,x2(t)=6c1e2t+c2e5t.

Our gallery Fig. 7.4.16 at the end of this section shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (8). (In Problem 29 we explore “Cartesian” equations for the solution curves (10) relative to the “axes” defined by the lines l1 and l2, which form a natural frame of reference for the solution curves.)

Nodes: Sinks and Sources

Distinct Negative Eigenvalues: When λ1<λ2<0, the factors eλ1t and eλ2t both decrease as t increases. Indeed, as t+, both eλ1t and eλ2t approach zero, which means that the solution curve

(4)x(t)=c1v1eλ1t+c2v2eλ2t

approaches the origin; likewise, as t, both eλ1t and eλ2t grow without bound, and so the solution curve “goes off to infinity.” Moreover, differentiation of the solution in (4) gives

(11)x(t)=c1λ1v1eλ1t+c2λ2v2eλ2t=eλ2t[c1λ1v1e(λ1λ2)t+c2λ2v2].

This shows that the tangent vector x(t) to the solution curve x(t) is a scalar multiple of the vector c1λ1v1e(λ1λ2)t+c2λ2v2, which approaches the fixed nonzero multiple c2λ2v2 of the vector v2 as t+ (because e(λ1λ2)t approaches zero). It follows that if c20, then as t+, the solution curve x(t) becomes more and more nearly parallel to the eigenvector v2. (More specifically, note that if c2>0, for example, then x(t) approaches the origin in the direction opposite to v2, because the scalar c2λ2 is negative.) Thus, if c20, then with increasing t the solution curve approaches the origin and is tangent there to the line l2 passing through the origin and parallel to v2.

If c2=0, on the other hand, then the solution curve x(t) flows similarly along the line l1 passing through the origin and parallel to theeigenvector v1. Once again, the net effect is that the lines l1 and l2 divide the plane into four “quadrants” as shown in Figure 7.4.2, which illustrates typical solution curves corresponding to nonzero values of the coefficients c1 and c2.

FIGURE 7.4.2.

Solution curves x(t)=c1v1eλ1t+c2v2eλ2t for the system x=Ax when the eigenvalues λ1,λ2 of A are real with λ1<λ2<0.

To describe the appearance of phase portraits like Fig. 7.4.2, we introduce some new terminology, which will be useful both now and in Chapter 9, when we study nonlinear systems. In general, we call the origin a node of the system x=Ax provided that both of the following conditions are satisfied:

Moreover, we say that the origin is a proper node provided that no two different pairs of “opposite” trajectories are tangent to the same straight line through the origin. This is the situation in Fig. 7.4.6, in which the trajectories are straight lines, not merely tangent to straight lines;indeed, a proper node might be called a “star point.” However, in Fig. 7.4.2, all trajectories—apart from those that flow along the line l1—are tangent to the line l2; as a result we call the node improper.

Further, if every trajectory for the system x=Ax approaches the origin as t+ (as in Fig. 7.4.2), then the origin is called a sink; if instead every trajectory recedes from the origin, then the origin is a source. Thus we describe the characteristic pattern of the trajectories in Fig. 7.4.2 as an improper nodal sink.

Example 2

The solution curves in Fig. 7.4.2 correspond to the choice

(12)A=[83213]

in the system x=Ax. The eigenvalues of A are λ1=14 and λ2=7 (and thus λ1<λ2<0), with associated eigenvectors

v1=[12]andv2=[31].

Equation (4) then gives the general solution

(13)x(t)=c1[12]e14t+c2[31]e7t,

or, in scalar form,

x1(t)=c1e14t+3c2e7t,x2(t)=2c1e14t+c2e7t.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (12).

The case of distinct positive eigenvalues mirrors that of distinct negative eigenvalues. But instead of analyzing it independently, we can rely on the following principle, whose verification is a routine matter of checking signs (Problem 30).

We note furthermore that the two vector-valued functions x(t) and x~(t) for <t< have the same solution curve (or image) in the plane. However, the chain rule gives x~(t)=x(t); since x~(t) and x(t) represent the same point, it follows that at each point of their common solution curve the velocity vectors of the two functions x(t) and x~(t) are negatives of each other. Therefore the two solutions traverse their common solution curve in opposite directions as t increases—or, alternatively, in the same direction as t increases for one solution and decreases for the other. In short, we may say that the solutions of the systems (1) and (14) correspond to each other under “time reversal,” since we get the solutions of one system by letting time “run backwards” in the solutions of the other.

Distinct Positive Eigenvalues: If the matrix A has positive eigenvalues with 0<λ2<λ1, then as you can verify (Problem 31), the matrix A has negative eigenvalues λ1<λ2<0 but the same eigenvectors v1 and v2. The preceding case then shows that the system x=Ax has an improper nodal sink at the origin. But the system x=Ax has the same trajectories, except with the direction of motion (as t increases) along each solution curve reversed. Thus the origin is now a source, rather than a sink, for the system x=Ax, and we call the origin an improper nodal source. Figure 7.4.3 illustrates typical solution curves given by x(t)=c1v1eλ1t+c2v2eλ2t corresponding to nonzero values of the coefficients c1 and c2.

FIGURE 7.4.3.

Solution curves x(t)=c1v1eλ1t+c2v2eλ2t for the system x=Ax when the eigenvalues λ1,λ2 of A are real with 0<λ2<λ1.

Example 3

The solution curves in Fig. 7.4.3 correspond to the choice

(15)A=[83213]=[83213]

in the system x=Ax; thus A is the negative of the matrix in Example 2. Therefore we can solve the system x=Ax by applying the principle of time reversal to the solution in Eq. (13): Replacing t with t in the righthand side of (13) leads to

(16)x(t)=c1[12]e14t+c2[31]e7t.

Of course, we could also have “started from scratch” by finding the eigenvalues λ1, λ2 and eigenvectors v1, v2 of A. These can be found from the definition of eigenvalue, but it is easier to note (see Problem 31 again)that because A is the negative of the matrix in Eq. (12), λ1 and λ2 are likewise the negatives of their values in Example 2, whereas we can take v1 and v2 to be the same as in Example 2. By either means we find that λ1=14 and λ2=7 (so that 0<λ2<λ1), with associated eigenvectors

v1=[12]andv2=[31].

From Eq. (4), then, the general solution is

x(t)=c1[12]e14t+c2[31]e7t

(in agreement with Eq. (16)), or, in scalar form,

x1(t)=c1e14t+3c2e7t,x2(t)=2c1e14t+c2e7t.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (15).

Zero Eigenvalues and Straight-Line Solutions

One Zero and One Negative Eigenvalue: When λ1<λ2=0, the general solution (4) becomes

(17)x(t)=c1v1eλ1t+c2v2.

For any fixed nonzero value of the coefficient c1, the term c1v1eλ1t in Eq. (17) is a scalar multiple of the eigenvector v1, and thus (as t varies) travels along the line l1 passing through the origin and parallel to v1; the direction of travel is toward the origin as t+ because λ1<0. If c1>0, for example, then c1v1eλ1t extends in the direction of v1, approaching the origin as t increases, and receding from the origin as t decreases. If instead c1<0, then c1v1eλ1t extends in the direction opposite v1 while still approaching the origin as t increases. Loosely speaking, we can visualize the flow of the term c1v1eλ1t taken alone as a pair of arrows opposing each other head-to-head at the origin. The solution curve x(t) in Eq. (17) is simply this same trajectory c1v1eλ1t, then shifted (or offset) by the constant vector c2v2. Thus in this case the phase portrait of the system x=Ax consists of all lines parallel to the eigenvector v1, where along each such line the solution flows (from both directions) toward the line l2 passing through the origin and parallel to v1. Figure 7.4.4 illustrates typical solution curves corresponding to nonzero values of the coefficients c1 and c2.

FIGURE 7.4.4.

Solution curves x(t)=c1v1eλ1t+c2v2eλ2t for the system x=Ax when the eigenvalues λ1,λ2 of A are real with λ1<λ2=0.

It is noteworthy that each single point represented by a constant vector b lying on the line l2 represents a constant solution of the system x=Ax. Indeed, if b lies on l2, then b is a scalar multiple k·v2 of the eigenvector v2 of A associated with the eigenvalue λ2=0. In this case, the constant-valued solution x(t)b is given by Eq. (17) with c1=0 and c2=k. This constant solution, with its “trajectory” being a single point lying on the line l2, is then the unique solution of the initial value problem

x=Ax,x(0)=b

guaranteed by Theorem 1 of Section 7.1. Note that this situation is in marked contrast with the other eigenvalue cases we have considered so far, in which x(t)0 is the only constant solution of the system x=Ax. (In Problem 32 we explore the general circumstances under which the system x=Ax has constant solutions other than x(t)0.)

Example 4

The solution curves in Fig. 7.4.4 correspond to the choice

(18)A=[36661]

in the system x=Ax. The eigenvalues of A are λ1=35 and λ2=0, with associated eigenvectors

v1=[61]andv2=[16].

Based on Eq. (17), the general solution is

(19)x(t)=c1[61]e35t+c2[16],

or, in scalar form,

x1(t)=6c1e35t+c2,x2(t)=c1e35t6c2.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (18).

One Zero and One Positive Eigenvalue: When 0=λ2<λ1, the solution of the system x=Ax is again given by

(17)x(t)=c1v1eλ1t+c2v2.

By the principle of time reversal, the trajectories of the system x=Ax are identical to those of the system x=Ax, except that they flow in the opposite direction. Since the eigenvalues λ1 and λ2 of the matrix A satisfy λ1<λ2=0, by the preceding case the trajectories of x=Ax are lines parallel to the eigenvector v1 and flowing toward the line l2 from both directions. Therefore the trajectories of the system x=Ax are lines parallel to v1 and flowing away from the line l2. Figure 7.4.5 illustrates typical solution curves given by x(t)=c1v1eλ1t+c2v2 corresponding to nonzero values of the coefficients c1 and c2.

FIGURE 7.4.5.

Solution curves x(t)=c1v1eλ1t+c2v2eλ2t for the system x=Ax when the eigenvalues λ1,λ2 of A are real with 0=λ2<λ1.

Example 5

The solution curves in Fig. 7.4.5 correspond to the choice

(20)A=[36661]=[36661]

in the system x=Ax; thus A is the negative of the matrix in Example 4. Once again we can solve the system using the principle of time reversal:Replacing t with t in the right-hand side of the solution in Eq. (19) of Example 4 leads to

(21)x(t)=c1[61]e35t+c2[16].

Alternatively, directly finding the eigenvalues and eigenvectors of A leads to λ1=35 and λ2=0, with associated eigenvectors

v1=[61]andv2=[16].

Equation (17) gives the general solution of the system x=Ax as

x(t)=c1[61]e35t+c2[16]

(in agreement with Eq. (21)), or, in scalar form,

x1(t)=6c1e35t+c2,x2(t)=c1e35t6c2.

Our gallery Fig.7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (20).

Repeated Eigenvalues; Proper and Improper Nodes

Repeated Positive Eigenvalue: As we noted earlier, if the matrix A has one repeated eigenvalue, then A may or may not have two associated linearly independent eigenvectors. Because these two possibilities lead to quite different phase portraits, we will consider them separately. We let λ denote the repeated eigenvalue of A with λ>0.

With two independent eigenvectors: First, if A does have two linearly independent eigenvectors, then it is easy to show (Problem 33) that in fact every nonzero vector is an eigenvector of A, from which it follows that A must be equal to the scalar λ times the identity matrix of order two, that is,

(22)A=λ[1001]=[λ00λ].

Therefore the system x=Ax becomes (in scalar form)

(23)x1(t)=λx1(t),x2(t)=λx2(t).

The general solution of Eq. (23) is

(24)x1(t)=c1eλtx2(t)=c2eλt,

or in vector format,

(25)x(t)=eλt[c1c2].

We could also have arrived at Eq. (25) by starting, as in previous cases, from our general solution (4): Because all nonzero vectors are eigenvectors of A, we are free to take v1=[10]T and v2=[01]T as a representative pair of linearly independent eigenvectors, each associated with the eigenvalue λ. Then Eq. (4) leads to the same result as Eq. (25):

x(t)=c1v1eλt+c2v2eλt=eλt(c1v1+c2v2)=eλt[c1c2].

Either way, our solution in Eq. (25) shows that x(t) is always a positive scalar multiple of the fixed vector [c1c2]T. Thus apart from the case c1=c2=0, the trajectories of the system (1) are half-lines, or rays, emanating from the origin and (because λ>0) flowing away from it. As noted above, the origin in this case represents a proper node, because no two pairs of “opposite” solution curves are tangent to the same straight line through the origin. Moreover the origin is also a source (rather than a sink), and so in this case we call the origin a proper nodal source. Figure 7.4.6 shows the “exploding star” pattern characteristic of such points.

Example 6

The solution curves in Fig.7.4.6 correspond to the case where the matrix A is given by Eq. (22) with λ=2:

(26)A=[2002].

FIGURE 7.4.6.

Solution curves x(t)=eλt[c1c2] for the system x=Ax when A has one repeated positive eigenvalue and two linearly independent eigenvectors.

Equation (25) then gives the general solution of the system x=Ax as

(27)x(t)=e2t[c1c2],

or, in scalar form,

x1(t)=c1e2t,x2(t)=c2e2t.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (26).

Without two independent eigenvectors: The remaining possibility is that the matrix A has a repeated positive eigenvalue yet fails to have two linearly independent eigenvectors. In this event the general solution of the system x=Ax is given by Eq. (7) above:

(7)x(t)=c1v1eλt+c2(v1t+v2)eλt.

Here v1 is an eigenvector of the matrix A associated with the repeated eigenvalue λ and v2 is a (nonzero) “generalized eigenvector” that will be described more fully in Section 7.6. To analyze this trajectory, we first distribute the factor eλt in Eq. (7), leading to

(28)x(t)=c1v1eλt+c2(v1teλt+v2eλt).

Our assumption that λ>0 implies that both eλt and teλt approach zero as t, and so by Eq. (28) the solution x(t) approaches the origin as t. Except for the trivial solution given by c1=c2=0, all trajectories given by Eq. (7) “emanate” from the origin as t increases.

The direction of flow of these curves can be understood from the tangent vector x(t). Rewriting Eq. (28) as

x(t)=eλt[c1v1+c2(v1t+v2)]

and applying the product rule for vector-valued functions gives

(29)x(t)=eλtc2v2+λeλt[c1v1+c2(v1t+v2)]=eλt(c2v1+λc1v1+λc2v1t+λc2v2).

For t0, we can factor out t in Eq. (29) and rearrange terms to get

(30)x(t)=teλt[λc2v1+1t(λc1v1+λc2v2+c2v1)].

Equation (30) shows that for t0, the tangent vector x(t) is a nonzero scalar multiple of the vector λc2v1+1t(λc1v1+λc2v2+c2v1), which, if c20, approaches the fixed nonzero multiple λc2v1 of the eigenvector v1 as t+ or as t. In this case it follows that as t gets larger and larger numerically (in either direction), the tangent line to the solution curve at the point x(t)—since it is parallel to the tangent vector x(t) which approaches λc2v1—becomes more and more nearly parallel to the eigenvector v1. In short, we might say that as t increases numerically, the point x(t) on the solution curve moves in a direction that is more and more nearly parallel to the vector v1, or still more briefly, that near x(t) the solution curve itself is virtually parallel to v1.

We conclude that if c20, then as t the point x(t) approaches the origin along the solution curve which is tangent there to the vector v1. But as t+ and the point x(t) recedes farther and farther from the origin, the tangent line to the trajectory at this point tends to differ (in direction) less and less from the (moving) line through x(t) that is parallel to the (fixed) vector v1. Speaking loosely but suggestively, we might therefore say that at points sufficiently far from the origin, all trajectories are essentially parallel to the single vector v1.

If instead c2=0, then our solution (7) becomes

(31)x(t)=c1v1eλt,

and thus runs along the line l1 passing through the origin and parallel to the eigenvector v1. Because λ>0, x(t) flows away from the origin as t increases; the flow is in the direction of v1 if c1>0, and opposite v1 if c1<0.

We can further see the influence of the coefficient c2 by writing Eq. (7) in yet a different way:

(32)x(t)=c1v1eλt+c2(v1t+v2)eλt=(c1+c2t)v1eλt+c2v2eλt.

It follows from Eq. (32) that if c20, then the solution curve x(t) does not cross the line l1. Indeed, if c2>0, then Eq. (32) shows that for all t, the solution curve x(t) lies on the same side of l1 as v2, whereas if c2<0, then x(t) lies on the opposite side of l1.

To see the overall picture, then, suppose for example that the coefficient c2>0. Starting from a large negative value of t, Eq. (30) shows that as t increases, the direction in which the solution curve x(t) initially proceeds from the origin is roughly that of the vector teλtλc2v1. Since the scalar teλtλc2 is negative (because t<0 and λc2>0), the direction of the trajectory is opposite that of v1. For large positive values of t, on the other hand, the scalar teλtλc2 is positive, and so x(t) flows in nearly the same direction as v1. Thus, as t increases from to +, the solution curve leaves the origin flowing in the direction opposite v1, makes a “U-turn” as it moves away from the origin, and ultimately flows in the direction of v1.

Because all nonzero trajectories are tangent at the origin to the line l1, the origin represents an improper nodal source. Figure 7.4.7 illustrates typical solution curves given by x(t)=c1v1eλt+c2(v1t+v2)eλt for the system x=Ax when A has a repeated eigenvalue but does not have two linearly independent eigenvectors.

FIGURE 7.4.7.

Solution curves x(t)=c1v1eλt+c2(v1t+v2)eλt for the system x=Ax when A has one repeated positive eigenvalue λ with associated eigenvector v1 and “generalized eigenvector” v2.

Example 7

The solution curves in Fig. 7.4.7 correspond to the choice

(33)A=[1337]

in the system x=Ax. In Examples 2 and 3 of Section 7.6 we will see that A has the repeated eigenvalue λ=4 with associated eigenvector and generalized eigenvector given by

(34)v1=[33]andv2=[10],

respectively. According to Eq. (7) the resulting general solution is

(35)x(t)=c1[33]e4t+c2[3t+13t]e4t,

or, in scalar form,

x1(t)=(3c2t3c1+c2)e4t,x2(t)=(3c2t+3c1)e4t.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (33).

Repeated Negative Eigenvalue: Once again the principle of time reversal shows that the solutions x(t) of the system x=Ax are identical to those of x=Ax with t replaced by t; hence these two systems share the same trajectories while flowing in opposite directions. Further, if the matrix A has the repeated negative eigenvalue λ, then the matrix A has the repeated positive eigenvalue λ (Problem 31 again). Therefore, to construct phase portraits for the system x=Ax when A has a repeated negative eigenvalue, we simply reverse the directions of the trajectories in the phase portraits corresponding to a repeated positive eigenvalue. These portraits are illustrated in Figs. 7.4.8 and 7.4.9. In Fig. 7.4.8 the origin represents a proper nodal sink, whereas in Fig. 7.4.9 it represents an improper nodal sink.

FIGURE 7.4.8.

Solution curves x(t)=eλt[c1c2] for the system x=Ax when A has one repeated negative eigenvalue λ and two linearly independent eigenvectors.

FIGURE 7.4.9.

Solution curves x(t)=c1v1eλt+c2(v1t+v2)eλt for the system x=Ax when A has one repeated negative eigenvalue λ with associated eigenvector v1 and “generalized eigenvector” v2.

Example 8

The solution curves in Fig. 7.4.8 correspond to the choice

(36)A=[2002]=[2002]

in the system x=Ax; thus A is the negative of the matrix in Example 6. We can solve this system by applying the principle of time reversal to the solution found in Eq. (27): Replacing t with t in the right-hand side of Eq. (27) leads to

(37)x(t)=e2t[c1c2],

or, in scalar form,

x1(t)=c1e2t,x1(t)=c2e2t.

Alternatively, because A is given by Eq. (22) with λ=2, Eq. (25) leads directly to the solution in Eq. (37). Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (36).

Example 9

The solution curves in Fig. 7.4.9 correspond to the choice

(38)A=[1337]=[1337]

in the system x=Ax. Thus A is the negative of the matrix in Example 7, and once again we can apply the principle of time reversal to the solution found in Eq. (35): Replacing t with t in the right-hand side of Eq. (35) yields

(39)x(t)=c1[33]e4t+c2[3t+13t]e4t.

We could also arrive at an equivalent form of the solution in Eq. (39) in the following way. You can verify that A has the repeated eigenvalue λ=2 with eigenvector v1 given by Eq. (34), that is,

v1=[33].

However, as the methods of Section 7.6 will show, a generalized eigenvector v2 associated with v1 is now given by

v2=[10]=[10];

that is, v2 is the negative of the generalized eigenvector in Eq. (34). Equation (7) then gives the general solution of the system x=Ax as

(40)x(t)=c1[33]e4t+c2[3t13t]e4t,

or, in scalar form,

x1(t)=(3c2t3c1c2)e4t,x2(t)=(3c2t+3c1)e4t.

Note that replacing c2 with c2 in the solution (39) yields the solution (40), thus confirming that the two solutions are indeed equivalent. Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (38).

The Special Case of a Repeated Zero Eigenvalue

Repeated Zero Eigenvalue: Once again the matrix A may or may not have two linearly independent eigenvectors associated with the repeated eigenvalue λ=0. If it does, then (using Problem 33 once more)we conclude that every nonzero vector is an eigenvector of A, that is, that Av=0·v=0 for all two-dimensional vectors v. It follows that A is the zero matrix of order two, that is,

A=[0000].

Therefore the system x=Ax reduces to x1(t)=x2(t)=0, which is to say that x1(t) and x2(t) are each constant functions. Thus the general solution of x=Ax is simply

(41)x(t)=[c1c2],

where c1 and c2 are arbitrary constants, and the “trajectories” given by Eq. (41) are simply the fixed points (c1,c2) in the phase plane.

If instead A does not have two linearly independent eigenvectors associated with λ=0, then the general solution of the system x=Ax is given by Eq. (7) with λ=0:

(42)x(t)=c1v1+c2(v1t+v2)=(c1v1+c2v2)+c2v2t.

Once again v1 denotes an eigenvector of the matrix A associated with the repeated eigenvalue λ=0 and v2 denotes a corresponding nonzero “generalized eigenvector.” If c20, then the trajectories given by Eq. (42) are lines parallel to the eigenvector v1 and “starting” at the point c1v1+c2v2 (when t=0). When c2>0 the trajectory proceeds in the same direction as v1, whereas when c2<0 the solution curve flows in the direction opposite v1. Once again the lines l1 and l2 passing through the origin and parallel to the vectors v1 and v2, respectively, divide the plane into “quadrants” corresponding to the signs of the coefficients c1 and c2. The particular quadrant in which the “starting point” c1v1+c2v2 of the trajectory falls is determined by the signs of c1 and c2. Finally, if c2=0, then Eq. (42) gives x(t)c1v1 for all t, which means that each fixed point c1v1 along the line l1 corresponds to a solution curve. (Thus the line l1 could be thought of as a median strip dividing two opposing lanes of traffic.) Figure 7.4.10 illustrates typical solution curves corresponding to nonzero values of the coefficients c1 and c2.

FIGURE 7.4.10.

Solution curves x(t)=(c1v1+c2v2)+c2v1t for the system x=Ax when A has a repeated zero eigenvalue with associated eigenvector v1 and “generalized eigenvector” v2. The emphasized point on each solution curve corresponds to t=0.

Example 10

The solution curves in Fig. 7.4.10 correspond to the choice

(43)A=[2412]

in the system x=Ax. You can verify that v1=[21]T is an eigenvector of A associated with the repeated eigenvalue λ=0. Further, using the methods of Section 7.6 we can show that v2=[10]T is a corresponding “generalized eigenvector” of A. According to Eq. (42) the general solution of the system x=Ax is therefore

(44)x(t)=c1[21]+c2([21]t+[10]),

or, in scalar form,

x1(t)=2c1+(2t+1)c2,x2(t)=c1tc2.

Our gallery Fig.7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (43).

Complex Conjugate Eigenvalues and Eigenvectors

We turn now to the situation in which the eigenvalues λ1 and λ2 of the matrix A are complex conjugate. As we noted at the beginning of this section, the general solution of the system x=Ax is given by Eq. (5):

(5)x(t)=c1ept(acosqtbsinqt)+c2ept(bcosqt+asinqt).

Here the vectors a and b are the real and imaginary parts, respectively, of a (complex-valued) eigenvector of A associated with the eigenvalue λ1=p+iq. We will divide the case of complex conjugate eigenvalues according to whether the real part p of λ1 and λ2 is zero, positive, or negative:

Pure Imaginary Eigenvalues: Centers and Elliptical Orbits

Pure Imaginary Eigenvalues: Centers and Elliptical Orbits Here we assume that the eigenvalues of the matrix A are given by λ1 λ2=±iq with q0. Taking p=0 in Eq. (5) gives the general solution

(45)x(t)=c1(acosqtbsinqt)+c2(bcosqt+asinqt).

for the system x=Ax. Rather than directly analyze the trajectories given by Eq. (45), as we have done in the previous cases, we begin instead with an example that will shed light on the nature of these solution curves.

Example 11

Solve the initial value problem

(46)x=[617816]x,x(0)=[42].

Solution

The coefficient matrix

(47)A=[61786]

has characteristic equation

|AλI|=[6λ1786λ]=λ2+100=0,

and hence has the complex conjugate eigenvalues λ1, λ2=±10i. If v=[ab]T is an eigenvector associated with λ=10i, then the eigenvector equation (AλI)v=0 yields

[A10iI]v=[610i178610i][ab]=[00].

Upon division of the second row by 2, this gives the two scalar equations

(48)(610i)a17b=0,4a(3+5i)b=0,

each of which is satisfied by a=3+5i and b=4. Thus the desired eigenvector is v=[3+5i4]T, with real and imaginary parts

(49)a=[34]andb=[50],

respectively. Taking q=10 in Eq. (45) therefore gives the general solution of the system x=Ax:

(50)x(t)=c1([34] cos 10t[50] sin 10t)+([50] cos 10t+[34] sin 10t)=[c1(3 cos 10t5 sin 10t)+c2(5 cos 10t+3 sin 10t)4c1 cos 10t+4c2 sin 10t].

To solve the given initial value problem it remains only to determine values of the coefficients c1 and c2. The initial condition x(0)=[42]T readily yields c1=c2=12, and with these values Eq. (50) becomes (in scalar form)

(51)x1(t)=4 cos 10tsin 10t,x2(t)=2 cos 10t+2 sin 10t.

Figure 7.4.11 shows the trajectory given by Eq. (51) together with the initial point (4, 2).

FIGURE 7.4.11.

Solution curve x1(t)=4cos10tsin10t,x2(t)=2cos10t+2sin10t for the initial value problem in Eq. (46).

This solution curve appears to be an ellipse rotated counterclockwise by the angle θ=arctan240.4636. We can verify this by finding the equations of the solution curve relative to the rotated u- and v-axes shown in Fig. 7.4.11. By a standard formula from analytic geometry, these new equations are given by

(52)u=x1cosθ+x2sinθ=25x1+15x2,v=x1sinθ+x2cosθ=15x1+25x2.

In Problem 34 we ask you to substitute the expressions for x1 and x2 from Eq. (51) into Eq. (52), leading (after simplification) to

(53)u=25 cos 10t,v=5 sin 10t.

Equation (53) not only confirms that the solution curve in Eq. (51) is indeed an ellipse rotated by the angle θ, but it also shows that the lengths of the semi-major and semi-minor axes of the ellipse are 25 and 5, respectively.

Furthermore, we can demonstrate that any choice of initial point (apart from the origin) leads to a solution curve that is an ellipse rotated by the same angle θ and “concentric” (in an obvious sense) with the trajectory in Fig.7.4.11 (see Problems 35–37). All these concentric rotated ellipses are centered at the origin (0, 0), which is therefore called a center for the system x=Ax whose coefficient matrix A has pure imaginary eigenvalues. Our gallery Fig.7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (47).

Further investigation: Geometric significance of the eigenvector. Our general solution in Eq. (50) was based upon the vectors a and b in Eq. (49), that is, the real and imaginary parts of the complex eigenvector v=[3+5i4]T of the matrix A. We might therefore expect a and b to have some clear geometric connection to the solution curve in Fig. 7.4.11. For example, we might guess that a and b would be parallel to the major and minor axes of the elliptical trajectory. However, it is clear from Fig. 7.4.12—which shows the vectors a and b together with the solution curve given by Eq. (51)—that this is not the case. Do the eigenvectors of A, then, play any geometric role in the phase portrait of the system x=Ax?

FIGURE 7.4.12.

Solution curve for the initial value problem in Eq. (46) showing the vectors a, b, a˜, and b˜.

The (affirmative) answer lies in the fact that any nonzero real or complex multiple of a complex eigenvector of the matrix A is still an eigenvector of A associated with that eigenvalue. Perhaps, then, if we multiply the eigenvector v=[3+5i4]T by a suitable nonzero complex constant z, the resulting eigenvector v~ will have real and imaginary parts a~ and b~ that can be readily identified with geometric features of the ellipse. To this end, let us multiply v by the complex scalar z=12(1+i). (The reason for this particular choice will become clear shortly.)The resulting new complex eigenvector v~ of the matrix A is,

v˜=zv=12(1+i)[3+5i4]=[1+4i2+2i],

and has real and imaginary parts

a˜=[12]andb˜=[42].

It is clear that the vector b~ is parallel to the major axis of our elliptical trajectory. Further, you can easily check that a~·b~=0, which means that a~ is perpendicular to b~, and hence is parallel to the minor axis of the ellipse, as Fig. 7.4.12 illustrates. Moreover, the length of b~ is twice that of a~, reflecting the fact that the lengths of the major and minor axes of the ellipse are in this same ratio. Thus for a matrix A with pure imaginary eigenvalues, the complex eigenvector of A used in the general solution (45)—if suitably chosen—is indeed of great significance to the geometry of the elliptical solution curves of the system x=Ax.

How was the value 12(1+i) chosen for the scalar z? In order that the real and imaginary parts a~ and b~ of v~=z·v be parallel to the axes of the ellipse, at a minimum a~ and b~ must be perpendicular to each other. In Problem 38 we ask you to show that this condition is satisfied if and only if z is of the form r(1±i), where r is a nonzero real number, and that if z is chosen in this way, then a~ and b~ are in fact parallel to the axes of the ellipse. The value r=12 then aligns the lengths of a~ and b~ with those of the semi-minor and -major axes of the elliptical trajectory. More generally, we can show that given any eigenvector v of a matrix A with pure imaginary eigenvalues, there exists a constant z such that the real and imaginary parts a~ and b~ of the eigenvector v~=z·v are parallel to the axes of the (elliptical) trajectories of the system x=Ax.

Further investigation: Direction of flow. Figs. 7.4.11 and 7.4.12 suggest that the solution curve in Eq. (51) flows in a counterclockwise direction with increasing t. However, you can check that the matrix

A=[61786]

has the same eigenvalues and eigenvectors as the matrix A in Eq. (47) itself, and yet (by the principle of time reversal) the trajectories of the system x=Ax are identical to those of x=Ax while flowing in the opposite direction, that is, clockwise. Clearly, mere knowledge of the eigenvalues and eigenvectors of the matrix A is not sufficient to predict the direction of flow of the elliptical trajectories of the system x=Ax as t increases. How then can we determine this direction of flow?

One simple approach is to use the tangent vector x to monitor the direction in which the solution curves flow as they cross the positive x1-axis. If s is any positive number (so that the point (s, 0) lies on the positive x1-axis), and if the matrix A is given by

A=[abcd],

then any trajectory for the system x=Ax passing through (s, 0) satisfies

x=Ax=[abcd][s0]=[ascs]=s[ac]

at the point (s, 0). Therefore, at this point the direction of flow of the solution curve is a positive scalar multiple of the vector [ac]T. Since c cannot be zero (see Problem 39), this vector either points “upward” into the first quadrant of the phase plane (if c>0 ), or “downward” into the fourth quadrant (if c<0). If upward, then the flow of the solution curve is counterclockwise; if downward, then clockwise. For the matrix A in Eq. (47), the vector [ac]T=[68]T points into the first quadrant because c=8>0, thus indicating a counterclockwise direction of flow (as Figs. 7.4.11 and 7.4.12 suggest).

Complex Eigenvalues: Spiral Sinks and Sources

Complex Eigenvalues with Negative Real Part: Now we assume that the eigenvalues of the matrix A are given by λ1, λ2=p±iq with q0 and p<0. In this case the general solution of the system x=Ax is given directly by Eq. (5):

(5)x(t)=c1ept(a cos qtb sinqt)+c2ept(b cos qt+a sin qt),

where the vectors a and b have their usual meaning. Once again we begin with an example to gain an understanding of these solution curves.

Example 12

Solve the initial value problem

(54)x=[51787]x,x(0)=[42].

Solution

The coefficient matrix

(55)A=[51787]

has characteristic equation

|AλI|=|5λ1787λ|=(λ+1)2+100=0,

and hence has the complex conjugate eigenvalues λ1, λ2=1±10i. If v=[ab]T is an eigenvector associated with λ=1+10i, then the eigenvector equation (AλI)v=0 yields the same system (48) of equations found in Example 11:

(48)(610i)a17b=0,4a(3+5i)b=0.

As in Example 11, each of these equations is satisfied by a=3+5i and b=4. Thus the desired eigenvector, associated with λ1=1+10i, is once again v=[3+5i4]T, with real and imaginary parts

(56)a=[34]andb=[50],

respectively. Taking p=1 and q=10 in Eq. (5) therefore gives the general solution of the system x=Ax:

(57)x(t)=c1et([34]cos 10t[50]sin 10t)+c2et([50]cos 10t+[34]sin 10t)=[c1et(3 cos 10t5 sin 10t)+c2et(5 cos 10t+3 sin 10t)4c1et cos 10t+4c2et sin 10t].

The initial condition x(0)=[42]T gives c1=c2=12 once again, and with these values Eq. (57) becomes (in scalar form)

(58)x1(t)=et(4cos10tsin10t),x2(t)=et(2cos10t+2sin10t).

Figure 7.4.13 shows the trajectory given by Eq. (58) together with the initial point (4, 2). It is noteworthy to compare this spiral trajectory with the elliptical trajectory in Eq. (51). The equations for x1(t) and x2(t) in (58) are obtained by multiplying their counterparts in (51) by the common factor et, which is positive and decreasing with increasing t. Thus for positive values of t, the spiral trajectory is generated, so to speak, by standing at the origin and “reeling in” the point on the elliptical trajectory (51) as it is traced out. When t is negative, the picture is rather one of “casting away” the point on the ellipse farther out from the origin to create the corresponding point on the spiral.

FIGURE 7.4.13.

Solution curve x1(t)=et(4 cos 10t sin 10t), x2(t)=et(2 cos 10t+2 sin 10t). for the initial value problem in Eq. (54). The dashed and solid portions of the curve correspond to negative and positive values of t, respectively.

Our gallery Fig. 7.4.16 shows a more complete set of solution curves, together with a direction field, for the system x=Ax with A given by Eq. (55). Because the solution curves all “spiral into” the origin, we call the origin in this a case a spiral sink.

Complex Eigenvalues with Positive Real Part: We conclude with the case where the eigenvalues of the matrix A are given by λ1, λ2=p±iq with q0 and p>0. Just as in the preceding case, the general solution of the system x=Ax is given by Eq. (5):

(5)x(t)=c1ept(a cos qtb sinqt)+c2ept(b cos qt+a sin qt).

An example will illustrate the close relation between the cases p>0 and p<0.

Example 13

Solve the initial value problem

(59)x=[51787]x,x(0)=[42].

Solution

Although we could directly apply the eigenvalue/eigenvector method as in previous cases (see Problem 40), here it is more convenient to notice that the coefficient matrix

(60)A=[51787]

is the negative of the matrix in Eq. (55) used in Example 12. By the principle of time reversal, therefore, the solution of the initial value problem (59) is given by simply replacing t with t in the right-hand sides of the solution (58) of the initial value problem in that example:

(61)x1(t)=et(4 cos 10t+sin 10t),x2(t)=et(2 cos 10t2 sin 10t).

Figure 7.4.14 shows the trajectory given by Eq. (61) together with the initial point (4, 2). Our gallery Fig. 7.4.16 shows this solution curve together with a direction field for the system x=Ax with A given by Eq. (60). Because the solution curve “spirals away from” the origin, we call the origin in this case a spiral source.

FIGURE 7.4.14.

Solution curve x1(t)=et(4 cos 10t+sin 10t), x2(t)=et(2 cos 10t2 sin 10t) for the initial value problem in Eq. (59). The dashed and solid portions of the curve correspond to negative and positive values of t, respectively.

A 3-Dimensional Example

Figure 7.4.15 illustrates the space trajectories of solutions of the 3-dimensional system x=Ax with constant coefficient matrix

(62)A=[4100560001].

To portray the motion in space of a point x(t) moving on a trajectory of this system, we can regard this trajectory as a necklace string on which colored beads are placed to mark its successive positions at fixed increments of time (so the point is moving fastest where the spacing between beads is greatest). In order to aid the eye in following the moving point’s progress, the size of the beads decreases continuously with the passage of time and motion along the trajectory.

FIGURE 7.4.15.

Three-dimensional trajectories for the system x=Ax with the matrix A given by Eq. (62).

The matrix A has the single real eigenvalue 1 with the single (real) eigenvector [001]T and the complex conjugate eigenvalues 1±5i. The negative real eigenvalue corresponds to trajectories that lie on the x3-axis and approach the origin as t0 (as illustrated by the beads on the vertical axis of the figure). Thus the origin (0, 0, 0) is a sink that “attracts” all the trajectories of the system.

The complex conjugate eigenvalues with negative real part correspond to trajectories in the horizontal x1x2-plane that spiral around the origin while approaching it. Any other trajectory—one which starts at a point lying neither on the z-axis nor in the x1x2-plane—combines the preceding behaviors by spiraling around the surface of a cone while approaching the origin at its vertex.

Gallery of Typical Phase Portraits for the System x=Ax: Nodes

FIGURE 7.4.16.

Gallery of typical phase plane portraits for the system x=Ax.

Proper Nodal Source: A repeated positive real eigenvalue with two linearly independent eigenvectors.

Proper Nodal Sink: A repeated negative real eigenvalue with two linearly independent eigenvectors.

Improper Nodal Source: Distinct positive real eigenvalues (left) or a repeated positive real eigenvalue without two linearly independent eigenvectors (right).

Improper Nodal Sink: Distinct negative real eigenvalues (left) or a repeated negative real eigenvalue without two linearly independent eigenvectors (right).

Gallery of Typical Phase Portraits for the System x=Ax: Saddles, Centers, Spirals, and Parallel Lines

FIGURE 7.4.16.

(Continued)

Saddle Point: Real eigenvalues of opposite sign.

Center: Pure imaginary eigenvalues.

Spiral Source: Complex conjugate eigenvalues with positive real part.

Spiral Sink: Complex conjugate eigenvalues with negative real part.

Parallel Lines: One zero and one negative real eigenvalue. (If the nonzero eigenvalue is positive, then the trajectories flow away from the dotted line.)

Parallel Lines: A repeated zero eigenvalue without two linearly independent eigenvectors.

7.4 Problems

For each of the systems in Problems 1 through 16 in Section 7.3, categorize the eigenvalues and eigenvectors of the coefficient matrix A according to Fig. 7.4.16 and sketch the phase portrait of the system by hand. Then use a computer system or graphing calculator to check your answer.

The phase portraits in Problems 17 through 28 correspond to linear systems of the form x=Ax in which the matrix A has two linearly independent eigenvectors. Determine the nature of the eigenvalues and eigenvectors of each system. For example, you may discern that the system has pure imaginary eigenvalues, or that it has real eigenvalues of opposite sign; that an eigenvector associated with the positive eigenvalue is roughly [21]T, etc.

  1. We can give a simpler description of the general solution

    (9)x(t)=c1[16]e2t+c2[11]e5t

    of the system

    x=[4161]x

    in Example 1 by introducing the oblique uv-coordinate system indicated in Fig. 7.4.17, in which the u- and v-axes are determined by the eigenvectors v1=[16] and v2=[11], respectively.

    FIGURE 7.4.17.

    The oblique uv-coordinate system determined by the eigenvectors v1 and v2.

    The uv-coordinate functions u(t) and v(t) of the moving point x(t) are simply its distances from the origin measured in the directions parallel to v1 and v2. It follows from (9) that a trajectory of the system is described by

    (63)u(t)=u0e2t,v(t)=v0e5t

    where u0=u(0) and v0=v(0). (a) Show that if v0=0, then this trajectory lies on the u-axis, whereas if u0=0, then it lies on the v-axis. (b) Show that if u0 and v0 are both nonzero, then a “Cartesian” equation of the parametric curve in Eq. (63) is given by v=Cu5/2.

  2. Use the chain rule for vector-valued functions to verify the principle of time reversal.

In Problems 31–33 A represents a 2×2 matrix.

  1. Use the definitions of eigenvalue and eigenvector (Section 7.3) to prove that if λ is an eigenvalue of A with associated eigenvector v, then λ is an eigenvalue of the matrix A with associated eigenvector v. Conclude that if A has positive eigenvalues 0<λ2<λ1 with associated eigenvectors v1 and v2, then A has negative eigenvalues λ1<λ2<0 with the same associated eigenvectors.

  2. Show that the system x=Ax has constant solutions other than x(t)0 if and only if there exists a (constant) vector x0 with Ax=0. (It is shown in linear algebra that such a vector x exists exactly when det(A)=0.)

  3. (a) Show that if A has the repeated eigenvalue λ with two linearly independent associated eigenvectors, then every nonzero vector v is an eigenvector of A. (Hint: Express v as a linear combination of the linearly independent eigenvectors and multiply both sides by A.) (b) Conclude that A must be given by Eq. (22). (Suggestion: In the equation Av=λv take v=[10]Tand v=[01]T.)

  4. Verify Eq. (53) by substituting the expressions for x1(t) and x2(t) from Eq. (51) into Eq. (52) and simplifying.

Problems 35–37 show that all nontrivial solution curves of the system in Example 11 are ellipses rotated by the same angle as the trajectory in Fig. 7.4.11.

  1. The system in Example 11 can be rewritten in scalar form as

    x1=6x117x2,x2=8x16x2,

    leading to the first-order differential equation

    dx2dx1=dx2/dtdx1/dt=8x16x26x117x2,

    or, in differential form,

    (6x28x1)dx1+(6x117x2)dx2=0.

    Verify that this equation is exact with general solution

    (64)4x12+6x1x2172x22=k,

    where k is a constant.

  2. In analytic geometry it is shown that the general quadratic equation

    (65)Ax12+Bx1x2+Cx22=k

    represents an ellipse centered at the origin if and only if Ak>0 and the discriminant B24AC<0. Show that Eq. (64) satisfies these conditions if k<0, and thus conclude that all nondegenerate solution curves of the system in Example 11 are elliptical.

  3. It can be further shown that Eq. (65) represents in general a conic section rotated by the angle θ given by

    tan 2θ=BAC.

    Show that this formula applied to Eq. (64) leads to the angle θ=arctan24 found in Example 11, and thus conclude that all elliptical solution curves of the system in Example 11 are rotated by the same angle θ. (Suggestion: You may find useful the double-angle formula for the tangent function.)

  4. Let v=[3+5i4]T be the complex eigenvector found in Example 11 and let z be a complex number. (a) Show that the real and imaginary parts a~ and b~, respectively, of the vector v~=z·v are perpendicular if and only if z=r(1±i) for some nonzero real number r. (b) Show that if this is the case, then a~ and b~ are parallel to the axes of the elliptical trajectory found in Example 11 (as Fig. 7.4.12 indicates).

  5. Let A denote the 2×2 matrix

    A=[abcd].
    1. Show that the characteristic equation of A (Eq. (3), Section 6.1) is given by

      λ2(a+d)λ+(adbc)=0.
    2. Suppose that the eigenvalues of A are pure imaginary. Show that the trace T(A)=a+d of A must be zero and that the determinant D(A)=adbc must be positive. Conclude that c0.

  6. Use the eigenvalue/eigenvector method to confirm the solution in Eq. (61) of the initial value problem in Eq. (59).

7.4 Application Dynamic Phase Plane Graphics

Using computer systems we can “bring to life” the static gallery of phase portraits in Fig. 7.4.16 by allowing initial conditions, eigenvalues, and even eigenvectors to vary in “real time.” Such dynamic phase plane graphics afford additional insight into the relationship between the algebraic properties of the 2×2 matrix A and the phase plane portrait of the system x=Ax.

For example, the basic linear system

dx1dt=x1,dx2dt=kx2(k a nonzero constant),

has general solution

x1(t)=aet,x2(t)=bekt,

where (a, b) is the initial point. If a0, then we can write

(1)x2=bekt=bak(ae1)=cx1k,

where c=b/ak. A version of the Maple commands

with(plots):
createPlot := proc(k)
   soln := plot([exp(-t), exp(-k*t),
      t = -10..10], x = -5..5, y = -5..5):
return display(soln):
end proc:
Explore(createPlot(k),
   parameters = [k = -2.0..2.0])

produces Fig. 7.4.18, which allows the user to vary the parameter k continuously from k=2 to k=2, thus showing dynamically the changes in the solution curves (1) in response to changes in k.

FIGURE 7.4.18.

Interactive display of the solution curves in Eq. (1). Using the slider, the value of k can be varied continuously from 2 to 2.

Figure 7.4.19 shows snapshots of the interactive display in Fig. 7.4.18 corresponding to the values 1,12, and 2 for the parameter k. Based on this progression, how would you expect the solution curves in Eq. (1) to look when k=1? Does Eq. (1) corroborate your guess?

FIGURE 7.4.19.

Snapshots of the interactive display in Fig. 7.4.18 with the initial conditions held fixed and the parameter k equal to 1,12, and 2, respectively.

As another example, a version of the Mathematica commands

a = {{-5, 17}, {-8, 7}};
x[t_] := {x1[t], x2[t]};
Manipulate[
   soln = DSolve[{x′[t] == a.x[t],
      x[0] == pt[[1]]}, x[t], t];
   ParametricPlot[x[t]/.soln, {t, -3.5, 10},
      PlotRange -> 5],
   {{pt, {{4, 2}}}, Locator}]

was used to generate Fig. 7.4.20, which (like Figs. 7.4.13 and 7.4.14) shows the solution curve of the initial value problem

(2)x=[51787]x,x(0)=[42]

from Example 13 of the preceding section. However, in Fig. 7.4.20 the initial condition (4, 2) is attached to a “locator point” which can be freely dragged to any desired position in the phase plane, with the corresponding solution curve being instantly redrawn—thus illustrating dynamically the effect of varying the initial conditions.

FIGURE 7.4.20.

Interactive display of the initial value problem in Eq. (2). As the “locator point” is dragged to different positions, the solution curve is immediately redrawn, showing the effect of changing the initial conditions.

FIGURE 7.4.21.

Interactive display of the initial value problem x=Ax with A given by Eq. (3). Both the initial conditions and the value of the parameter k can be varied dynamically.

Finally, Fig. 7.4.21 shows a more sophisticated, yet perhaps more revealing, demonstration. As you can verify, the matrix

(3)A=110[k+933k33k9k+1]

has the variable eigenvalues 1 and k but with fixed associated eigenvectors [31]T and [13]T, respectively. Figure 7.4.21, which was generated by a version of the Mathematica commands

a[k ] := (1/10){{k + 9, 3 - 3k}, {3 - 3k, 9k + 1}}
x[t ] := {x1[t], x2[t]}
Manipulate[
   soln[k ] = DSolve[{x′[t] == a[k].x[t],
      x[0] == #}, x[t], t]&/@pt;
   curve = ParametricPlot
      [Evaluate[x[t]/.soln[k]], {t, -10, 10},
      PlotRange -> 4], {k, -1, 1},
   {{pt, {{2, -1}, {1, 2}, {-1, -2}, {-2, 1}}},
   Locator}]

shows the phase portrait of the system x=Ax with A given by Eq. (3). Not only are the initial conditions of the individual trajectories controlled independently by the “locator points,” but using the slider we can also vary the value of k continuously from 1 to 1, with the solution curves being instantly redrawn. Thus for a fixed value of k we can experiment with changing initial conditions throughout the phase plane, or, conversely, we can hold the initial conditions fixed and observe the effect of changing the value of k.

As a further example of what such a display can reveal, Fig. 7.4.22 consists of a series of snapshots of Fig. 7.4.21 where the initial conditions are held fixed and k progresses through the specific values 1, 0.25, 0, 0.5, 0.65, and 1. The result is a “video” showing stages in a transition from a saddle point with “hyperbolic” trajectories, to a pair of parallel lines, to an improper nodal source with “parabolic” trajectories, and finally to the exploding star pattern of a proper nodal source with straight-line trajectories. Perhaps these frames provide a new interpretation of the description “dynamical system” for a collection of interdependent differential equations.

FIGURE 7.4.22.

Snapshots of the interactive display in Fig. 7.4.21 with the initial conditions held fixed and the parameter k increasing from 1 to 1.

7.5 Second-Order Systems and Mechanical Applications*

In this section we apply the matrix methods of Sections 7.2 and 7.3 to investigate the oscillations of typical mass-and-spring systems having two or more degrees of freedom. Our examples are chosen to illustrate phenomena that are generally characteristic of complex mechanical systems.

Figure 7.5.1 shows three masses connected to each other and to two walls by the four indicated springs. We assume that the masses slide without friction and that each spring obeys Hooke’s law—its extension or compression x and force F of reaction are related by the formula F=kx. We further assume that when the system is at equilibrium (so the masses are stationary) none of the springs is stretched or compressed. If the rightward displacements x1,x2, and x3 of the three masses (from their respective equilibrium positions) are all positive, then

FIGURE 7.5.1.

Three spring-coupled masses.

Therefore, application of Newton’s law F=ma to the three masses (as in Example 1 of Section 7.1) yields their equations of motion:

(1)m1x1=k1x1+k2(x2x1),m2x2=k2(x2x1)+k3(x3x2),m3x3=k3(x3x2)k4x3.

Although we assumed in writing these equations that the displacements of themasses are all positive, they actually follow similarly from Hooke’s and Newton’s laws, whatever the signs of these displacements.

In terms of the displacement vector x=[x1x2x3]T, the mass matrix

(2)M=[m1000m2000m3]

and the stiffness matrix

(3)K=[(k1+k2)k20k2(k2+k3)k30k3(k3+k4)],

the system in (1) takes the matrix form

(4)Mx=Kx.

The notation in Eqs. (1) through (4) generalizes in a natural way to the system of n spring-coupled masses shown in Fig. 7.5.2. We need only write

(5)M=[m1000m2000mn]

FIGURE 7.5.2.

A system of n spring-coupled masses.

and

(6)K=[(k1+k2)k200k2(k2+k3)k300k3(k3+k4)000k4000(kn1+kn)kn00kn(kn+kn+1)]

for the mass and stiffness matrices in Eq. (4).

The diagonal matrix M is obviously nonsingular; to get its inverse M1 we need only replace each diagonal element with its reciprocal. Hence multiplication of each side in Eq. (4) by M1 yields the homogeneous second-order system

(7)x=Ax,

where A=M1K. There is a wide variety of frictionless mechanical systems for which a displacement or position vector x, a nonsingular mass matrix M, and a stiffness matrix K satisfying Eq. (4) can be defined.

Solution of Second-Order Systems

To seek a solution of Eq. (7), we substitute (as in Section 7.3 for a first-order system) a trial solution of the form

(8)x(t)=veαt,

where v is a constant vector. Then x=α2veαt, so substitution of Eq. (8) in (7) gives

α2veαt=Aveαt,

which implies that

(9)Av=α2v.

Therefore x(t)=veαt is a solution of x=Ax if and only if α2=λ, an eigenvalue of the matrix A, and v is an associated eigenvector.

If x=Ax models a mechanical system, then it is typical that the eigenvalues of A are negative real numbers. If

α2=λ=ω2<0,

then α=±ωi. In this case the solution given by Eq. (8) is

x(t)=veiωt=v(cos ωt+i sin ωt).

The real and imaginary parts

(10)x1(t)=v cos ωtandx2(t)=v sin ωt

of x(t) are then linearly independent real-valued solutions of the system. This analysis leads to the following theorem.

Remark

The nonzero vector v0 is an eigenvector corresponding to λ0=0 provided that Av0=0. If x(t)=(a0+b0t)v0, then

x=0v0=(a0+b0t)0=(a0+b0t)(Av0)=Ax,

thus verifying the form in Eq. (12).

Example 1

Mass-spring system Consider the mass-and-spring system with n=2 shown in Fig. 7.5.3. Because there is no third spring connected to a right-hand wall, we set k3=0. If m1=2,m2=1,k1=100, and k2=50, then the equation Mx=Kx is

(13)[2001]x=[150505050]x,

which reduces to x=Ax with

A=[75255050].

FIGURE 7.5.3.

The mass-and-spring system of Example 1.

The characteristic equation of A is

(75λ)(50λ)5025=λ2+125λ+2500=(λ+25)(λ+100)=0,

so A has the negative eigenvalues λ1=25 and λ2=100. By Theorem 1, the system in (13) therefore has solutions with [circular] frequencies ω1=5 and ω2=10.

Case 1: λ1=25. The eigenvector equation (AλI)v=0 is

[50255025][ab]=[00],

so an eigenvector associated with λ1=25 is v1=[12]T.

Case 2: λ2=100. The eigenvector equation (AλI)v=0 is

[25255050][ab]=[00],

so an eigenvector associated with λ2=100 is v2=[11]T.

By Eq. (11) it follows that a general solution of the system in (13) is given by

(14)x(t)=(a1 cos 5t+b1 sin 5t)v1+(a2 cos 10t+b2 sin 10t)v2.

The two terms on the right in Eq. (14) represent free oscillations of the mass-and-spring system. They describe the physical system’s two natural modes of oscillation at its two [circular] natural frequencies ω1=5 and ω2=10. The natural mode

x1(t)=(a1 cos 5t+b1 sin 5t)v1=c1 cos(5tα1)[12]

(with c1=a12+b12,cosα1=a1/c1, and sinα1=b1/c1) has the scalar component equations

(15)x1(t)=c1cos(5tα1),x2(t)=2c1cos(5tα1),

and therefore describes a free oscillation in which the two masses move in synchrony in the same direction and with the same frequency ω1=5, but with the amplitude of motion of m2 twice that of m1 (see Fig. 7.5.4). The natural mode

x2(t)=(a2 cos 10t+b2 sin 10t)v2=c2 cos(10tα2)[11]

FIGURE 7.5.4.

Oscillations in the same direction with frequency ω1=5; the amplitude of motion of mass 2 is twice that of mass 1.

has the scalar component equations

(16)x1(t)=c2 cos(10tα2),x2(t)=c2 cos(10tα2),

and therefore describes a free oscillation in which the two masses move in synchrony in opposite directions with the same frequency ω2=10 and with equal amplitudes of oscillation (see Fig. 7.5.5).

FIGURE 7.5.5.

Oscillations in opposite directions with frequency ω2=10; the amplitudes of motion of the two masses are the same.

Example 2

Railway cars Figure 7.5.6 shows three railway cars connected by buffer springs that react when compressed, but disengage instead of stretching. With n=3,k2=k3=k, and k1=k4=0 in Eqs. (2) through (4), we get the system

(17)[m1000m2000m3]x=[kk0k2kk0kk]x,

which is equivalent to

(18)x=[c1c10c22c2c20c3c3]x

with

(19)ci=kmi(i=1,2,3).

FIGURE 7.5.6.

The three railway cars of Example 2.

If we assume further that m1=m3, so that c1=c3, then a brief computation gives

(20)λ(λ+c1)(λ+c1+2c2)=0

for the characteristic equation of the coefficient matrix A in Eq. (18). Hence the matrix A has eigenvalues

(21a)λ1=0,λ2=c1,λ3=c12c2

corresponding to the natural frequencies

(21b)ω1=0,ω2=c1,ω3=c1+2c2

of the physical system.

For a numerical example, suppose that the first and third railway cars weigh 12 tons each, that the middle car weighs 8 tons, and that the spring constant is k=1:5 tons/ft; i.e., k=3000 lb/ft. Then, using fps units with mass measured in slugs (a weight of 32 pounds has a mass of 1 slug), we have

m1=m3=750,m2=500,

and

c1=3000750=4,c2=3000500=6.

Hence the coefficient matrix A is

(22)A=[4406126044],

and the eigenvalue-frequency pairs given by (21a) and (21b) are λ1=0,ω1=0;λ2=4,ω2=2; and λ3=16,ω3=4.

Case 1: λ1=0,ω1=0. The eigenvector equation (AλI)v=0 is

(22)Av=[4406126044] [abc]=[000],

so it is clear that v1=[111]T is an eigenvector associated with λ1=0. According to Theorem 1, the corresponding part of a general solution of x=Ax is

x1(t)=(a1+b1t)v1.

Case 2: λ2=4,ω2=2. The eigenvector equation (AλI)v=0 is

(A+4I)v=[040686040] [abc]=[000],

so it is clear that v2=[101]T is an eigenvector associated with λ2=4. According to Theorem 1, the corresponding part of a general solution of x=Ax is

x2(t)=(a2cos 2t+b2sin 2t)v2.

Case 3: λ3=16,ω3=4. The eigenvector equation (AλI)v=0 is

(A+16I)v=[12406460412] [abc]=[000],

so it is clear that v3=[131]T is an eigenvector associated with λ3=16. According to Theorem 1, the corresponding part of a general solution of x=Ax is

x3(t)=(a3cos4t+b3sin4t)v3.

The general solution x=x1+x2+x3 of x=Ax is therefore given by

(23)x(t)=a1[111]+b1t[111]+a2[101]cos2t+b2[101]sin2t+a3[131]cos4t+b3[131]sin4t.

To determine a particular solution, let us suppose that the leftmost car is moving to the right with velocity v0 and at time t=0 strikes the other two cars, which are together but at rest. The corresponding initial conditions are

(24a)x1(0)=x2(0)=x3(0)=0,
(24b)x1(0)=v0,x2(0)=x3(0)=0.

Then substitution of (24a) in (23) gives the scalar equations

a1+a2+a3=0,a13a3=0,a1a2+a3=0,

which readily yield a1=a2=a3=0. Hence the position functions of the three cars are

(25)x1(t)=b1t+b2sin 2t+b3sin 4t,x2(t)=b1t3b3sin 4t,x3(t)=b1tb2sin 2t+b3sin 4t,

and their velocity functions are

(26)x1!(t)=b1+2b2 cos 2t+4b3 cos 4t,x2!(t)=b112b3 cos 4t,x3!(t)=b12b2 cos 2t+4b3 cos 4t.

Substitution of (24b) in (26) gives the equations

b1+2b2+4b3=v0,b112b3=0,b12b2+4b3=0

that readily yield b1=38v0, b2=14v0, and b3=132v0. Finally, the position functions in are

(27)x1(t)=132v0(12t+8 sin 2t+ sin 4t),x2(t)=132v0(12t3 sin 4t),x3(t)=132v0(12t8 sin 2t+ sin 4t).

But these equations hold only so long as the two buffer springs remain compressed; that is, while both

x2x1<0andx3x2<0.

To discover what this implies about t, we compute

x2(t)x1(t)=132v0(8 sin 2t4 sin 4t)=132v0(8 sin 2t+8 sin 2tcos 2t)=14v0( sin 2t)(1+cos 2t)

and, similarly,

x3(t)x2(t)=14v0(sin 2t)(1cos 2t).

It follows that x2x1<0 and x3x2<0 until t=π/21.57 (seconds), at which time the equations in (26) and (27) give the values

x1(π2)=x2(π2)=x3(π2)=3πv016,x1(π2)=x2(π2)=0,x3(π2)=v0.

We conclude that the three railway cars remain engaged and moving to the right until disengagement occurs at time t=π/2. Thereafter, cars 1 and 2 remain at rest (!), while car 3 continues to the right with speed v0. If, for instance, v0=48 feet per second (about 33 miles per hour), then the three cars travel a distance of 9π28.27 (ft) during their 1:57 seconds of engagement, and

(27′)x1(t)=x2(t)=9π,x3(t)=48t15π

for t>π/2. Figure 7.5.7 illustrates the “before” and “after” situations, and Fig. 7.5.8 shows the graphs of the functions x1(t),x2(t), and x3(t) in Eqs. (27) and (27′).

FIGURE 7.5.7.

(a) Before; (b) after.

FIGURE 7.5.8.

Position functions of the three railway cars of Example 2.

Forced Oscillations and Resonance

Suppose now that the ith mass of the mass-and-spring system in Fig. 7.5.2 is subject to an external force Fi(i=1,2,,n) in addition to the forces exerted by the springs attached to it. Then the homogeneous equation Mx=Kx is replaced with the nonhomogeneous equation

(28)Mx=Kx+F,

where F=[F1F2Fn]T is the external force vector for the system. Multiplication by M1 yields

(29)x=Ax+f,

where f is the external force vector per unit mass. We are especially interested in the case of a periodic external force

(30)f(t)=F0 cos ωt

(where F0 is a constant vector). We then anticipate a periodic particular solution

(31)xp(t)=c cos wt

with the known external frequency ω and with a coefficient vector c yet to be determined. Because xp=ω2c cos ωt, substitution of (30) and (31) in (29), followed by cancellation of the common factor cos ωt, gives the linear system

(32)(A+ω2I)c=F0

to be solved for c.

Observe that the matrix A+ω2I is nonsingular—in which case Eq. (32) can be solved for c—unless ω2=λ, an eigenvalue of A. Thus a periodic particular solution of the form in Eq. (31) exists provided that the external forcing frequency does not equal one of the natural frequencies ω1,ω2,,ωn of the system. The case in which ω is a natural frequency corresponds to the phenomenon of resonance discussed in Section 5.6.

Example 3

Mass-spring resonance Suppose that the second mass in Example 1 is subjected to the external periodic force 50 cos ωt. Then with m1=2, m2=1, k1=100, k2=50, and F0=50 in Fig. 7.5.9, Eq. (29) takes the form

(33)x=[75255050] x+[050]cos ωt,

FIGURE 7.5.9.

The forced mass-and-spring system of Example 3.

and the substitution x=c cos ωt leads to the equation

(34)[ω2752550ω250]c=[050]

for the coefficient vector c=[c1c2]T. This system is readily solved for

(35)c1=1250(ω225)(ω2100),c2=50(ω275)(ω225)(ω2100).

For instance, if the external squared frequency is ω2=50, then (35) yields c1=1, c2=1. The resulting forced periodic oscillation is described by

x1(t)=cos ωt,x2(t)=cos ωt.

Thus the two masses oscillate in synchrony with equal amplitudes and in the same direction.

If the external squared frequency is ω2=125, then (35) yields c1=12, c2=1. The resulting forced periodic oscillation is described by

x1(t)=12cos ωt,x2(t)=cos ωt.

and now the two masses oscillate in synchrony in opposite directions, but with the amplitude of motion of m2 twice that of m1.

It is evident from the denominators in (35) that c1 and c2 approach + as ω approaches either of the two natural frequencies ω1=5 and ω2=10 (found in Example 1). Figure 7.5.10 shows a plot of the amplitude c12+c22 of the forced periodic solution x(t)=c cos ωt as a function of the forced frequency ω. The peaks at ω2=5 and ω2=10 exhibit visually the phenomenon of resonance.

FIGURE 7.5.10.

Frequency–amplitude plot for Example 3.

Periodic and Transient Solutions

It follows from Theorem 4 of Section 7.2 that a particular solution of the forced system

(36)x=Ax+F0 cos ωt

will be of the form

(37)x(t)=xc(t)+xp(t),

where xp(t) is a particular solution of the nonhomogeneous system and xc(t) is a solution of the corresponding homogeneous system. It is typical for the effects of frictional resistance in mechanical systems to damp out the complementary function solution xc(t), so that

(38)xc(t)0ast+.

Hence xc(t) is a transient solution that depends only on the initial conditions; it dies out with time, leaving the steady periodic solution xp(t) resulting from the external driving force:

(39)x(t)xp(t)ast+.

As a practical matter, every physical system includes frictional resistance (however small) that damps out transient solutions in this manner.

7.5 Problems

Problems 1 through 7 deal with the mass-and-spring system shown in Fig. 7.5.11 with stiffness matrix

K=[(k1+k2)k2k2(k2+k3)]

and with the given mks values for the masses and spring constants. Find the two natural frequencies of the system and describe its two natural modes of oscillation.

FIGURE 7.5.11.

The mass-and-spring system for Problems 1 through 6.

  1. m1=m2=1; k1=0, k2=2, k3=0 (no walls)

  2. m1=m2=1; k1=1, k2=4, k3=1

  3. m1=1, m2=2; k1=1, k2=k3=2

  4. m1=m2=1; k1=1, k2=2, k3=1

  5. m1=m2=1; k1=2, k2=1, k3=2

  6. m1=1, m2=2; k1=2, k2=k3=4

  7. m1=m2=1; k1=4, k2=6, k3=4

In Problems 8 through 10 the indicated mass-and-spring system is set in motion from rest (x1(0)=x2(0)=0) in its equilibrium position (x1(0)=x2(0)=0) with the given external forces F1(t) and F2(t) acting on the masses m1 and m2, respectively. Find the resulting motion of the system and describe it as a superposition of oscillations at three different frequencies.

  1. The mass-and-spring system of Problem 2, with F1(t)=96 cos 5t, F2(t)0

  2. The mass-and-spring system of Problem 3, with F1(t)0, F2(t)=120 cos 3t

  3. The mass-and-spring system of Problem 7, with F1(t)=30 cos t, F2(t)=60 cos t

  4. Consider a mass-and-spring system containing two masses m1=1 and m2=1 whose displacement functions x(t) and y(t) satisfy the differential equations

    x=40x+8y,y=12x60y.

    (a) Describe the two fundamental modes of free oscillation of the system. (b) Assume that the two masses start in motion with the initial conditions

    x(0)=19,x(0)=12

    and

    y(0)=3,y(0)=6

    and are acted on by the same force, F1(t)=F2(t)=195 cos 7t. Describe the resulting motion as a superposition of oscillations at three different frequencies.

In Problems 12 and 13, find the natural frequencies of the three-mass system of Fig. 7.5.1, using the given masses and spring constants. For each natural frequency ω, give the ratio a1:a2:a3 of amplitudes for a corresponding natural mode x1=a1cos ωt, x2=a2cos ωt, x3=a3cos ωt.

  1. m1=m2=m3=1; k1=k2=k3=k4=1

  2. m1=m2=m3=1; k1=k2=k3=k4=2 (Hint: One eigenvalue is λ=4.)

  3. In the system of Fig. 7.5.12, assume that m1=1, k1=50, k2=10, and F0=5 in mks units, and that ω=10. Then find m2 so that in the resulting steady periodic oscillations, the mass m1 will remain at rest(!). Thus the effect of the second mass-and-spring pair will be to neutralize the effect of the force on the first mass. This is an example of a dynamic damper. It has an electrical analogy that some cable companies use to prevent your reception of certain cable channels.

    FIGURE 7.5.12.

    The mechanical system of Problem 14.

  4. Suppose that m1=2, m2=12, k1=75, k2=25, F0=100, and ω=10 (all in mks units) in the forced mass-and-spring system of Fig. 7.5.9. Find the solution of the system Mx=Kx+F that satisfies the initial conditions x(0)=x(0)=0.

In Problems 16 through 19 we apply the analysis of Example 2 to a system of two railway cars.

  1. Figure 7.5.13 shows two railway cars with a buffer spring. We want to investigate the transfer of momentum that occurs after car 1 with initial velocity v0 impacts car 2 at rest. The analog of Eq. (18) in the text is

    x=[c1c1c2c2] x

    with ci=k/mi for i=1, 2. Show that the eigenvalues of the coefficient matrix A are λ1=0 and λ2=c1c2, with associated eigenvectors v1=[11]T and v2=[c1c2]T.

    FIGURE 7.5.13.

    The two railway cars of Problems 16 through 19.

  2. If the two cars of Problem 16 both weigh 16 tons (so that m1=m2=1000 (slugs)) and k=1 ton/ft (that is, 2000 lb/ft), show that the cars separate after π/2 seconds, and that x1(t)=0 and x2(t)=v0 thereafter. Thus the original momentum of car 1 is completely transferred to car 2.

  3. If cars 1 and 2 weigh 8 and 16 tons, respectively, and k=3000 lb/ft, show that the two cars separate after π/3 seconds, and that

    x1(t)=13v0andx2(t)=+23v0

    thereafter. Thus the two cars rebound in opposite directions.

  4. If cars 1 and 2 weigh 24 and 8 tons, respectively, and k=1500 lb/ft, show that the cars separate after π/2 seconds, and that

    x1(t)=+12v0andx2(t)=+32v0

    thereafter. Thus both cars continue in the original direction of motion, but with different velocities.

Problems 20 through 23 deal with the same system of three railway cars (same masses) and two buffer springs (same spring constants) as shown in Fig. 7.5.6 and discussed in Example 2. The cars engage at time t=0 with x1(0)=x2(0)=x3(0)=0 and with the given initial velocities (where v0=48ft/s). Show that the railway cars remain engaged until t=π/2 (s), after which time they proceed in their respective ways with constant velocities. Determine the values of these constant final velocities x1(t), x2(t), and x3(t) of the three cars for t>π/2. In each problem you should find (as in Example 2) that the first and third railway cars exchange behaviors in some appropriate sense.

  1. x1(0)=v0, x2(0)=0, x3(0)=v0

  2. x1(0)=2v0, x2(0)=0, x3(0)=v0

  3. x1(0)=v0, x2(0)=v0, x3(0)=2v0

  4. x1(0)=3v0, x2(0)=2v0, x3(0)=2v0

  5. In the three-railway-car system of Fig. 7.5.6, suppose that cars 1 and 3 each weigh 32 tons, that car 2 weighs 8 tons, and that each spring constant is 4 tons/ft. If x1(0)=v0 and x2(0)=x3(0)=0, show that the two springs are compressed until t=π/2 and that

    x1(t)=19v0andx2(t)=x3(t)=+89v0

    thereafter. Thus car 1 rebounds, but cars 2 and 3 continue with the same velocity.

The Two-Axle Automobile

In Example 4 of Section 5.6 we investigated the vertical oscillations of a one-axle car—actually a unicycle. Now we can analyze a more realistic model: a car with two axles and with separate front and rear suspension systems. Figure 7.5.14 represents the suspension system of such a car. We assume that the car body acts as would a solid bar of mass m and length L=L1+L2. It has moment of inertia I about its center of mass C, which is at distance L1 from the front of the car. The car has front and back suspension springs with Hooke’s constants k1 and k2, respectively. When the car is in motion, let x(t) denote the vertical displacement of the center of mass of the car from equilibrium; let θ(t) denote its angular displacement (in radians) from the horizontal. Then Newton’s laws of motion for linear and angular acceleration can be used to derive the equations

(40)mx=(k1+k2)x+(k1L1k2L2)θ,I θ=(k1L1k2L2)x(k1L12k2L22)θ.

FIGURE 7.5.14.

Model of the two-axle automobile.

  1. Suppose that m=75 slugs (the car weighs 2400 lb), L1=7 ft, L2=3 ft (it’s a rear-engine car), k1=k2=2000 lb/ft, and I=1000 ft·lb·s2. Then the equations in (40) take the form

    75x+4000x8000θ=0,1000θ8000x+116,000θ=0.

    (a) Find the two natural frequencies ω1 and ω2 of the car. (b) Now suppose that the car is driven at a speed of v feet per second along a washboard surface shaped like a sine curve with a wavelength of 40 ft. The result is a periodic force on the car with frequency ω=2πv/40=πv/20. Resonance occurs when ω=ω1 or ω=ω2. Find the corresponding two critical speeds of the car (in feet per second and in miles per hour).

  2. Suppose that k1=k2=k and L1=L2=12L in Fig. 7.5.14 (the symmetric situation). Then show that every free oscillation is a combination of a vertical oscillation with frequency

    ω1=2k/m

    and an angular oscillation with frequency

    ω2=kL2/(2I).

In Problems 27 through 29, the system of Fig. 7.5.14 is taken as a model for an undamped car with the given parameters in fps units. (a) Find the two natural frequencies of oscillation (in hertz). (b) Assume that this car is driven along a sinusoidal washboard surface with a wavelength of 40 ft. Find the two critical speeds.

  1. m=100, I=800, L1=L2=5, k1=k2=2000

  2. m=100, I=1000, L1=6, L2=4, k1=k2=2000

  3. m=100, I=800, L1=L2=5, k1=1000, k2=2000

7.5 Application Earthquake-Induced Vibrations of Multistory Buildings

In this application you are to investigate the response to transverse earthquake ground oscillations of the seven-story building illustrated in Fig. 7.5.15. Suppose that each of the seven (above-ground) floors weighs 16 tons, so the mass of each is m=1000 (slugs). Also assume a horizontal restoring force of k=5 (tons per foot) between adjacent floors. That is, the internal forces in response to horizontal displacements of the individual floors are those shown in Fig. 7.5.16. It follows that the free transverse oscillations indicated in Fig. 7.5.15 satisfy the equation Mx=Kx with n=7, mi=1000 (for each i), and ki=10,000 (lb/ft) for 1i7. The system then reduces to the form x=Ax with

(1)A=[20100000010201000000102010000001020100000010201000000102010000001010].

FIGURE 7.5.15.

The seven-story building.

Once the matrix A has been entered, the TI-Nspire command eigVl(A) instantly computes the seven eigenvalues shown in the λ-column of the table in Fig. 7.5.17. Alternatively, you can use the Maple command eigenvals(A), the Matlab command eig(A), or the Mathematica command Eigenvalues[A].

FIGURE 7.5.16.

Forces on the ith floor.

FIGURE 7.5.17.

Frequencies and periods of natural oscillations of the seven-story building.

  Eigenvalue Frequency Period
i λ ω=λ P=2πω(sec)
1 38.2709 6.1863 1.0157
2 33.3826 5.7778 1.0875
3 26.1803 5.1167 1.2280
4 17.9094 4.2320 1.4847
5 10.0000 3.1623 1.9869
6 3.8197 1.9544 3.2149
7 0.4370 0.6611 9.5042

Then calculate the entries in the remaining columns of the table showing the natural frequencies and periods of oscillation of the seven-story building. Note that a typical earthquake producing ground oscillations with a period of 2 seconds is uncomfortably close to the fifth natural frequency (with period 1.9869 seconds) of the building.

A horizontal earthquake oscillation Ecos ωt of the ground, with amplitude E and acceleration a=Eω2 cos ωt, produces an opposite inertial force F=ma=mEω2 cos ωt on each floor of the building. The resulting nonhomogeneous system is

(2)x=Ax+(Eω2 cos ωt)b,

where b=[1111111]T and A is the matrix of Eq. (1). Figure 7.5.18 shows a plot of maximal amplitude (for the forced oscillations of any single floor) versus the period of the earthquake vibrations. The spikes correspond to the first six of the seven resonant frequencies. We see, for instance, that whereas an earthquake with period 2 (s) likely would produce destructive resonance vibrations in the building, it probably would be unharmed by an earthquake with period 2.5 (s). Different buildings have different natural frequencies of vibration, and so a given earthquake may demolish one building but leave untouched the one next door. This seeming anomaly was observed in Mexico City after the devastating earthquake of September 19, 1985.

For your personal seven-story building to investigate, let the weight (in tons) of each story equal the largest digit of your student ID number and let k (in tons/ft) equal the smallest nonzero digit. Produce numerical and graphical results like those illustrated in Figs. 7.5.17 and 7.5.18. Is your building susceptible to likely damage from an earthquake with period in the 2- to 3-second range?

FIGURE 7.5.18.

Resonance vibrations of a seven-story building—maximal amplitude as a function of period.

You might like to begin by working manually the following warm-up problems.

  1. Find the periods of the natural vibrations of a building with two above-ground floors, each weighing 16 tons and with each restoring force being k=5 tons/ft.

  2. Find the periods of the natural vibrations of a building with three above-ground floors, with each weighing 16 tons and with each restoring force being k=5 tons/ft.

  3. Find the natural frequencies and natural modes of vibration of a building with three above-ground floors as in Problem 2, except that the upper two floors weigh 8 tons each instead of 16. Give the ratios of the amplitudes A, B, and C of oscillations of the three floors in the form A:B:C with A=1.

  4. Suppose that the building of Problem 3 is subject to an earthquake in which the ground undergoes horizontal sinusoidal oscillations with a period of 3 s and an amplitude of 3 in. Find the amplitudes of the resulting steady periodic oscillations of the three above-ground floors. Assume the fact that a motion Esin ωt of the ground, with acceleration a=Eω2 sin ωt, produces an opposite inertial force F=ma=mEω2 sin ωt on a floor of mass m.

7.6 Multiple Eigenvalue Solutions

In Section 7.3 we saw that if the n×n matrix A has n distinct (real or complex) eigenvalues λ1, λ2, , λn with respective associated eigenvectors v1, v2, , vn, then a general solution of the system

(1)dxdt=Ax

is given by

(2)x(t)=c1v1eλ1t+c2v2eλ2t++cnvneλnt

with arbitrary constants c1, c2, , cn. In this section we discuss the situation when the characteristic equation

(3)|AλI|=0

does not have n distinct roots, and thus has at least one repeated root.

An eigenvalue is of multiplicity k if it is a k-fold root of Eq. (3). For each eigenvalue λ, the eigenvector equation

(4)|AλI|v=0

has at least one nonzero solution v, so there is at least one eigenvector associated with λ. But an eigenvalue of multiplicity k>1 may have fewer than k linearly independent associated eigenvectors. In this case we are unable to find a “complete set” of n linearly independent eigenvectors of A, as needed to form the general solution in (2).

Let us call an eigenvalue of multiplicity k complete if it has k linearly independent associated eigenvectors. If every eigenvalue of the matrix A is complete, then—because eigenvectors associated with different eigenvalues are linearly independent—it follows that A does have a complete set of n linearly independent eigenvectors v1, v2, , vn associated with the eigenvalues λ1, λ2, , λn (each repeated with its multiplicity). In this case a general solution of x=Ax is still given by the usual combination in (2).

Example 1

Find a general solution of the system

(5)x=[940610643]x.

Solution

The characteristic equation of the coefficient matrix A in Eq. (5) is

|AλI|=|9λ4061λ0643λ|=(3λ)[(9λ)(1λ)+24]=(3λ)(158λ+λ2)=(5λ)(3λ)2=0.

Thus A has the distinct eigenvalue λ1=5 and the repeated eigenvalue λ2=3 of multiplicity k=2.

Case 1: λ1=5. The eigenvector equation (AλI)v=0, where v=[abc]T, is

(A5I)v=[440660642][abc]=[000].

Each of the first two equations, 4a+4b=0 and 6a6b=0, yields b=a. Then the third equation reduces to 2a2c=0, so that c=a. The choice a=1 then yields the eigenvector

v1=[111]T

associated with the eigenvalue λ1=5.

Case 2: λ2=3. Now the eigenvector equation is

(A3I)v=[640640640][000].

so the nonzero vector v=[abc]T is an eigenvector if and only if

(6)6a+4b=0;

that is, b=32a. The fact that Eq. (6) does not involve c means that c is arbitrary, subject to the condition v0. If c=1, then we may choose a=b=0; this gives the eigenvector

v2=[001]T

associated with λ2=3. If c=0, then we must choose a to be nonzero. For instance, if a=2 (to avoid fractions), then b=3, so

v3=[230]T

is a second linearly independent eigenvector associated with the multiplicity 2 eigenvalue λ2=3.

Thus we have found a complete set v1, v2, v3 of three eigenvectors associated with the eigenvalues 5, 3, 3. The corresponding general solution of Eq. (5) is

(7)x(t)=c1v1e5t+c2v2e3t+c3v3e3t=c1[111]e5t+c2[001]e3t+c3[230]e3t,

with scalar component functions given by

x1(t)=c1e5t+2c3e3t,x2(t)=c1e5t3c3e3t,x3(t)=c1e5t+c2e3t.

Remark

Our choice in Example 1 of the two eigenvectors

v2=[001]Tandv2=[230]T

associated with the repeated eigenvalue λ2=3 bears comment. The fact that b=32a for any eigenvector associated with λ2=3 means that any such eigenvector can be written as

v=[a32ac]=c[001]+12a[230]=cv2+12av3,

and thus is a linear combination of v2 and v3. Therefore, given a and c not both zero, we could choose v rather than v3 as our third eigenvector, and the new general solution

x(t)=c1v1e5t+c2v2e3t+c3ve3t

would be equivalent to the one in Eq. (7). Thus we need not worry about making the “right” choice of independent eigenvectors associated with a multiple eigenvalue. Any choice will do; we generally make the simplest one we can.

Defective Eigenvalues

The following example shows that—unfortunately—not all multiple eigenvalues are complete.

Example 2

The matrix

(8)A=[1337]

has characteristic equation

|AλI|=|1λ337λ|=(1λ)(7λ)+9=λ28λ+16=(λ4)2=0.

Thus A has the single eigenvalue λ1=4 of multiplicity 2. The eigenvector equation

(A4I)v=[3333][ab]=[00]

then amounts to the equivalent scalar equations

3a3b=0,3a+3b=0.

Hence b=a if v=[ab]T is to be an eigenvector of A. Therefore, any eigenvector associated with λ1=4 is a nonzero multiple of v=[11]T. Thus the multiplicity 2 eigenvalue λ1=4 has only one independent eigenvector, and hence is incomplete.

An eigenvalue λ of multiplicity k>1 is called defective if it is not complete. If λ has only p<k linearly independent eigenvectors, then the number

(9)d=kp

of “missing” eigenvectors is called the defect of the defective eigenvalue λ. Thus the defective eigenvalue λ1=4 in Example 2 has multiplicity k=2 and defect d=1, because we saw that it has only p=1 associated eigenvector.

If the eigenvalues of the n×n matrix A are not all complete, then the eigenvalue method as yet described will produce fewer than the needed n linearly independent solutions of the system x=Ax. We therefore need to discover how to find the “missing solutions” corresponding to a defective eigenvalue λ of multiplicity k>1.

The Case of Multiplicity k=2

Let us begin with the case k=2, and suppose that we have found (as in Example 2) that there is only a single eigenvector v1 associated with the defective eigenvalue λ. Then at this point we have found only the single solution

(10)x1(t)=v1eλt

of x=Ax. By analogy with the case of a repeated characteristic root for a single linear differential equation (Section 5.3), we might hope to find a second solution of the form

(11)x1(t)=(v2t)eλt=v2teλt.

When we substitute x=v2teλt in x=Ax, we get the equation

v2eλt+λv2teλt=Av2teλt.

But because the coefficients of both eλt and teλt must balance, it follows that v2=0, and hence that x2(t)0. This means that—contrary to our hope—the system x=Ax does not have a nontrivial solution of the form assumed in (11).

Instead of simply giving up on the idea behind Eq. (11), let us extend it slightly and replace v2t with v1t+v2. Thus we explore the possibility of a second solution of the form

(12)x2(t)=(v1t+v2)eλt=v1eλt+v2eλt,

where v1 and v2 are nonzero constant vectors. When we substitute x=v1teλt+v2eλt in x=Ax, we get the equation

(13)v1eλt+λv1teλt+λv2eλt=Av1teλt+Av2eλt.

We equate coefficients of eλt and teλt here, and thereby obtain the two equations

(14)(AλI)v1=0

and

(15)(AλI)v2=v1

that the vectors v1 and v2 must satisfy in order for (12) to give a solution of x=Ax.

Note that Eq. (14) merely confirms that v1 is an eigenvector of A associated with the eigenvalue λ. Then Eq. (15) says that the vector v2 satisfies the equation

(AλI)2v2=(AλI)[(AλI)v2]=(AλI)v1=0.

It follows that, in order to solve simultaneously the two equations in (14) and (15), it suffices to find a solution v2 of the single equation (AλI)2v2=0 such that the resulting vector v1=(AλI)v2 is nonzero. It turns out that this is always possible if the defective eigenvalue λ of A is of multiplicity 2. Consequently, the procedure described in the following algorithm always succeeds in finding two independent solutions associated with such an eigenvalue.

Example 3

Find a general solution of the system

(20)x=[1337]x.

Solution

In Example 2 we found that the coefficient matrix A in Eq. (20) has the defective eigenvalue λ=4 of multiplicity 2. We therefore begin by calculating

(A4I)2=[3333][3333]=[0000].

Hence Eq. (16) is

[0000]v2=0.

and therefore is satisfied by any choice of v2. In principle, it could happen that (A4I)v2 is nonzero (as desired) for some choices of v2 though not for others. If we try v2=[10]T we find that

(A4I)v2=[3333][10]=[33]=v1

is nonzero, and therefore is an eigenvector associated with λ=4. (It is 3 times the eigenvector found in Example 2.) Therefore, the two solutions of Eq. (20) given by Eqs. (18) and (19) are

x1(t)=v1e4t=[33]e4t,x2(t)=(v1t+v2)e4t=[3t+13t]e4t.

The resulting general solution

x2(t)=c1x1(t)+c2x2(t)

has scalar component functions

x1(t)=(3c2t+c23c1)e4t,x2(t)=(3c2t+3c1)e4t.

With c2=0 these solution equations reduce to the equations x1(t)=3c1e4t, x2(t)=3c1e4t, which parametrize the line x1=x2 in the x1x2-plane. The point (x1(t),x2(t)) then recedes along this line away from the origin as t+, to the northwest if c1>0 and to the southeast if c1<0. As indicated in Fig. 7.6.1, each solution curve with c20 is tangent to the line x1=x2 at the origin; the point (x1(t),x2(t)) approaches the origin as t and approaches + along the solution curve as t+.

FIGURE 7.6.1.

Direction field and solution curves for the linear system x1=x13x2, x2=3x1+7x2 of Example 3.

Generalized Eigenvectors

The vector v2 in Eq. (16) is an example of a generalized eigenvector. If λ is an eigenvalue of the matrix A, then a rank r generalized eigenvector associated with λ is a vector v such that

(21)(AλI)v=0but(AλI)r1v0

If r=1, then (21) simply means that v is an eigenvector associated with λ (recalling the convention that the 0th power of a square matrix is the identity matrix). Thus a rank 1 generalized eigenvector is an ordinary eigenvector. The vector v2 in (16) is a rank 2 generalized eigenvector (and not an ordinary eigenvector).

The multiplicity 2 method described earlier boils down to finding a pair {v1,v2} of generalized eigenvectors, one of rank 1 and one of rank 2, such that (AλI)v2=v1. Higher multiplicity methods involve longer “chains” of generalized eigenvectors. A length k chain of generalized eigenvectors based on the eigenvector v1 is a set {v1,v2,,vk} of k generalized eigenvectors such that

(22)(AλI)vk=vk1,(AλI)vk1=vk2,(AλI)v2=v1

Because v1 is an ordinary eigenvector, (AλI)v1=0. Therefore, it follows from (22) that

(23)(AλI)kvk=0.

If {v1,v2,v3} is a length 3 chain of generalized eigenvectors associated with the multiple eigenvalue λ of the matrix A, then it is easy to verify that three linearly independent solutions of x=Ax are given by

(24)x1(t)=v1eλt,x2(t)=(v1t+v2)eλt,x3(t)=(12v1t2+v2t+v3)eλt.

For instance, the equations in (22) give

Av3=v2+λv3,Av2=v1+λv2,Av3=λv1,

so

Ax3=[12Av1t2+Av2t+Av3]eλt=[12λv1t2+(v1+λv2)t+(v2+λv3)]eλt=(v1t+v2)eλt+λ(12v1t2+v2t+v3)eλt=x3.

Therefore, x3(t) in (24) does, indeed, define a solution of x=Ax.

Consequently, in order to “handle” a multiplicity 3 eigenvalue λ, it suffices to find a length 3 chain {v1,v2,v3} of generalized eigenvalues associated with λ. Looking at Eq. (23), we see that we need only find a solution v3 of

(AλI)3v3=0

such that the vectors

v2=(AλI)v3andv1=(AλI)v2

are both nonzero (although, as we will see, this is not always possible).

Example 4

Find three linearly independent solutions of the system

(25)x=[012537100]x.

Solution

The characteristic equation of the coefficient matrix in Eq. (25) is

|AλI|=[λ1253λ710λ]=1·[72·(3λ)]+(λ)[(λ)(3λ)+5]=λ33λ23λ1=(λ+1)3=0,

and thus A has the eigenvalue λ=1 of multiplicity 3. The eigenvector equation (AλI)v=0 for an eigenvector v=[abc]T is

(A+I)v=[112527101][abc]=[000].

The third row a+c=0 gives c=a; then the first row a+b+2c=0 gives b=a. Thus, to within a constant multiple, the eigenvalue λ=1 has only the single associated eigenvector v=[aaa]T with a0, and so the defect of λ=1 is 2.

To apply the method described here for triple eigenvalues, we first calculate

(A+I)2=[112527101][112527101]=[213213213]

and

(A+I)3=[112527101][213213213]=[000000000].

Thus any nonzero vector v3 will be a solution of the equation (A+I)3v3=0. Beginning with v3=[100]T, for instance, we calculate

v2=(A+I)v3=[112527101][100]=[151],v1=(A+I)v2=[112527101][151]=[222].

Note that v1 is the previously found eigenvector v with a=2; this agreement serves as a check of the accuracy of our matrix computations.

Thus we have found a length 3 chain {v1,v2,v3} of generalized eigenvectors associated with the triple eigenvalue λ=1. Substitution in (24) now yields the linearly independent solutions

x1(t)=v1et=[222]et,x2(t)=(v1t+v2)et=[2t+12t52t+1]et,x3(t)=(12v1t2+v2t+v3)et=[t2+t+1t25tt2+t]et

of the system x=Ax.

The General Case

A fundamental theorem of linear algebra states that every n×n matrix A has n linearly independent generalized eigenvectors. These n generalized eigenvectors may be arranged in chains, with the sum of the lengths of the chains associated with a given eigenvalue λ equal to the multiplicity of λ. But the structure of these chains depends on the defect of λ, and can be quite complicated. For instance, a multiplicity 4 eigenvalue can correspond to

Note that, in each of these cases, the length of the longest chain is at most d+1, where d is the defect of the eigenvalue. Consequently, once we have found all the ordinary eigenvectors associated with a multiple eigenvalue λ, and therefore know the defect d of λ, we can begin with the equation

(26)(AλI)d+1u=0

to start building the chains of generalized eigenvectors associated with λ.

Each length k chain {v1,v2,,vk} of generalized eigenvectors (with v1 an ordinary eigenvector associated with λ) determines a set of k independent solutions of x=Ax corresponding to the eigenvalue λ:

(27)x1(t)=v1eλt,x2(t)=(v1t+v2)eλt,x3(t)=(12v1t+v2t+v3)eλt,xk(t)=(v1tk1(k1)!++vk2t22!+vk1t+vk)eλt.

Note that (27) reduces to Eqs. (18) through (19) and (24) in the cases k=2 and k=3, respectively.

To ensure that we obtain n generalized eigenvectors of the n×n matrix A that are actually linearly independent, and therefore produce a complete set of n linearly independent solutions of x=Ax when we amalgamate all the “chains of solutions” corresponding to different chains of generalized eigenvectors, we may rely on the following two facts:

Example 5

Suppose that the 6×6 matrix A has two multiplicity 3 eigenvalues λ1=2 and λ2=3 with defects 1 and 2, respectively. Then λ1 must have an associated eigenvector u1 and a length 2 chain {v1,v2} of generalized eigenvectors (with the eigenvectors u1 and v1 being linearly independent), whereas λ2 must have a length 3 chain {w1,w2,w3} of generalized eigenvectors based on its single eigenvector w1. The six generalized eigenvectors u1, v1, v2, w1, w2, and w3 are then linearly independent and yield the following six independent solutions of x=Ax:

x1(t)=u1e2t,x2(t)=v1e2t,x3(t)=(v1t+v2)e2t,x4(t)=w1e3t,x5(t)=(w1t+w2)e3t,x6(t)=(12w1t2+w2t+w3)e3t.

As Example 5 illustrates, the computation of independent solutions corresponding to different eigenvalues and chains of generalized eigenvalues is a routine matter. The determination of the chain structure associated with a given multiple eigenvalue can be more interesting (as in Example 6).

An Application

Figure 7.6.2 shows two railway cars that are connected with a spring (permanently attached to both cars) and with a damper that exerts opposite forces on the two cars, of magnitude c(x1x2) proportional to their relative velocity. The two cars are also subject to frictional resistance forces c1x1 and c2x2 proportional to their respective velocities. An application of Newton’s law ma=F (as in Example 1 of Section 7.1) yields the equations of motion

(28)m1x1=k(x2x1)c1x1c(x1x2),m2x2=k(x1x2)c2x2c(x2x1).

FIGURE 7.6.2.

The railway cars of Example 6.

In terms of the position vector x(t)=[x1(t)x2(t)]T, these equations can be written in the matrix form

(29)Mx=Kx+Rx.

where M and K are mass and stiffness matrices (as in Eqs. (2) and (3) of Section 7.4), and

R=[(c+c1)cc(c+c2)]

is the resistance matrix. Unfortunately, because of the presence of the term involving x, the methods of Section 7.4 cannot be used.

Instead, we write (28) as a first-order system in the four unknown functions x1(t), x2(t), x3(t)=x1(t), and x4(t)=x2(t). If m1=m2=1 we get

(30)x=Ax,

where now x=[x1x2x3x4]T and

(31)A=[00100001kk(c+c1)ckkc(c+c2)].

Example 6

Two railway cars With m1=m2=c=1 and k=c1=c2=2, the system in Eq. (30) is

(32)x=[0010000122312213]x.

It is not too tedious to calculate manually—although a computer algebra system such as Maple, Mathematica, or Matlab is useful here—the characteristic equation

λ4+6λ3+12λ2+8λ=λ(λ+2)2=0

of the coefficient matrix A in Eq. (32). Thus A has the distinct eigenvalue λ0=0 and the triple eigenvalue λ1=2.

Case 1: λ0=0. The eigenvalue equation (AλI)v=0 for the eigenvector v=[abcd]T is

Av=[0010000122312213][abcd]=[0000].

The first two rows give c=d=0; then the last two rows yield a=b. Thus

v0=[1100]T

is an eigenvector associated with λ0=0.

Case 2: λ1=2 The eigenvalue equation (AλI)v=0 is

(A+2I)v=[2010020122112211][abcd]=[0000].

The third and fourth scalar equations here are the differences of the first and second equations, and therefore are redundant. Hence v is determined by the first two equations,

2a+c=0and2b+d=0.

We can choose a and b independently, then solve for c and d. Thereby we obtain two eigenvectors associated with the triple eigenvalue λ1=2. The choice a=1, b=0 yields c=2, d=0 and thereby the eigenvector

u1=[1020]T.

The choice a=0, b=1 yields c=0, d=2 and thereby the eigenvector

u2=[1002]T.

Because λ1=2 has defect 1, we need a generalized eigenvector of rank 2, and hence a nonzero solution v2 of the equation

(A+2I)2v2=[2211221100000000]v2=0.

Obviously,

v2=[0011]T

is such a vector, and we find that

(A+2I)v2=[2010020122112211][0011]=[1122]=v1

is nonzero, and therefore is an eigenvector associated with λ1=2. Then {v1,v2} is the length 2 chain we need.

The eigenvector v1 just found is neither of the two eigenvectors u1 and u2 found previously, but we observe that v1=u1u2. For a length 1 chain w1 to complete the picture, we can choose any linear combination of u1 and u2 that is independent of v1. For instance, we could choose either w1=u1 or w1=u2. However, we will see momentarily that the particular choice

w1=u1+u2=[1122]T

yields a solution of the system that is of physical interest.

Finally, the chains {v0}, {w1}, and {v1,v2} yield the four independent solutions

(33)x1(t)=v0e0·t=[1100]T,x2(t)=w1e2t=[1122]T e2t,x3(t)=v1e2t=[1122]T e2t,x4(t)=(v1t+v2)e2t=[tt2t+12t1]T e2t

of the system x=Ax in (32).

The four scalar components of the general solution

x(t)=c1x1(t)+c2x2(t)+c3x3(t)+c4x4(t)

are described by the equations

(34)x1(t)=c1+e2t(c2+c3+c4t),x2(t)=c1+e2t(c2c3c4t),x2(t)=e2t(2c22c3+c42c4t),x2(t)=e2t(2c2+2c3c4+2c4t).

Recall that x1(t) and x2(t) are the position functions of the two masses, whereas x3(t)=x1(t) and x4(t)=x2(t) are their respective velocity functions.

For instance, suppose that x1(0)=x2(0)=0 and that x1(0)=x2(0)=v0. Then the equations

x1(0)=c1+c2+c3=0,x2(0)=c3+c2c3=0,x1(0)=2c22c3+c4=v0,x2(0)=2c2+2c3c4=v0

are readily solved for c1=12v0, c2=12v0, and c3=c4=0, so

x1(t)=x2(t)=12v0(1e2t),x1(t)=x2(t)=v0e2t.

In this case the two railway cars continue in the same direction with equal but exponentially damped velocities, approaching the displacements x1=x2=12v0 as t+.

It is of interest to interpret physically the individual generalized eigenvector solutions given in (33). The degenerate (λ0=0) solution

x1(t)=[1100]T

describes the two masses at rest with position functions x1(t)1 and x2(t)1. The solution

x2(t)=[1122]T e2t

corresponding to the carefully chosen eigenvector w1 describes damped motions x1(t)=e2t and x2(t)=e2t of the two masses, with equal velocities in the same direction. Finally, the solutions x3(t) and x4(t) resulting from the length 2 chain {v1,v2} both describe damped motion with the two masses moving in opposite directions.

The methods of this section apply to complex multiple eigenvalues just as to real multiple eigenvalues (although the necessary computations tend to be somewhat lengthy). Given a complex conjugate pair α±βi of eigenvalues of multiplicity k, we work with one of them (say, αβi) as if it were real to find k independent complex-valued solutions. The real and imaginary parts of these complex-valued solutions then provide 2k real-valued solutions associated with the two eigenvalues λ=αβi and λ̲=α+βi each of multiplicity k. See Problems 33 and 34.

The Jordan Normal Form

In Section 6.2 we saw that if the n×n matrix A has a complete set v1,v2,,vn of n linearly independent eigenvectors, then it is similar to a diagonal matrix. In particular,

(35)P1AP=D,

where P=[v1v2vn] and the diagonal elements of the diagonal matrix D are the corresponding eigenvalues λ1,λ2,,λn (not necessarily distinct). This result is a special case of the following general theorem, a proof of which can be found in Appendix B of Gilbert Strang, Linear Algebra and Its Applications (4th edition, Brooks Cole, 2006).

If the Jordan block Ji in (37) is of size k×k, then it corresponds to a length k chain of generalized eigenvectors based on the (ordinary) eigenvector vi. If all these generalized eigenvectors are arranged as column vectors in proper order corresponding to the appearance of the Jordan blocks in (36), the result is a nonsingular n×n matrix Q such that

(38)Q1AQ=J.

The block-diagonal matrix J such that A=QJQ1 is called the Jordan normal form of the matrix A and is unique [except for the order of appearance of the Jordan blocks in (36)].

Example 7

In Example 3 we saw that the matrix

A=[1337]

has the single eigenvalue λ=4 and the associated length 2 chain of generalized eigenvectors {v1,v2}, where v1=[33]T and v2=[10]T, based on the eigenvector v1=Av2. If we define

Q=[v1v2]=[3130],

then we find that the Jordan normal form of A is

J=Q1AQ=[4104].

Here J=J1 is a single 2×2 Jordan block corresponding to the single eigenvalue λ=4 of A.

Example 8

In Example 6 we saw that the matrix

A=[0010000122312213]

has the distinct eigenvalue λ1=0 corresponding to the eigenvector

v1=[1100]T

and the triple eigenvalue λ2=2 corresponding both to the eigenvector v2=[1020]T and to the length 2 chain of generalized eigenvectors {v3,v4} with v3=[1122]T and v4=[0011]T, such that v3=(A+2I)v4. If we define

Q=[v1v2v3v4]=[1110101002210021],

then (using a computer algebra system to ease the labor of calculation) we find that the Jordan normal form of A is

J=Q1AQ=[0000020000210002].

Here we see the 1×1 Jordan block J1=[0] corresponding to the eigenvalue λ1=0 of A, as well as the two Jordan blocks J2=[2] and J3=[2102] corresponding to the two linearly independent eigenvectors v2 and v3 associated with the eigenvalue λ2=2.

The General Cayley-Hamilton Theorem

In Section 6.3, we showed that every diagonalizable matrix A satisfies its characteristic equation p(λ)=|AλI|=0, that is, p(A)=0. We can now use the Jordan normal form to show that this is true whether or not A is diagonalizable.

If J=Q1AQ is the Jordan normal form of the matrix A, then p(A)=Qp(J)Q1, just as in the conclusion to the proof of Theorem 1 in Section 6.3. It therefore suffices to show that p(J)=0. If the Jordan blocks J1,J2,,Js in (36) have sizes k1,k2,,ks—that is, Ji is a ki×ki matrix—and corresponding eigenvalues λ1,λ2,,λs (respectively), then

p(λ)=(λ1λ)k1(λ2λ)k2(λsλ)ks,

and so

(39)p(J)=(λ1IJ)k1(λ2IJ)k2(λsIJ)ks.

Now p(J) has the same block-diagonal structure as J itself, and we see from (39) that the ith block of p(J) involves the factor

(λiIJi)ki=[0100001000010000]ki,

where the ki×ki matrix λiIJi has all elements zero except on its first superdiagonal. It is easily seen that (λiIJi)2 then has nonzero elements only on its second superdiagonal; (λiIJi)3 has nonzero elements only on its third superdiagonal; and so on, in turn, until we see that (λiIJi)ki=0. Hence it follows from (39) that the ith block of p(J) is the ki×ki zero matrix. This being true for each i=1,2,,s, we conclude that p(J)=0, as desired. This completes the proof of the general case of the Cayley-Hamilton theorem.

7.6 Problems

Find general solutions of the systems in Problems 1 through 22. In Problems 1 through 6, use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system.

  1. x=[2114]x

  2. x=[3111]x

  3. x=[1225]x

  4. x=[3115]x

  5. x=[7143]x

  6. x=[1449]x

  7. x=[200797002]x

  8. x=[2512018506613]x

  9. x=[1912840508433]x

  10. x=[13404882324003]x

  11. x=[304111101]x

  12. x=[101011111]x

  13. x=[101014013]x

  14. x=[001515412]x

  15. x=[290140131]x

  16. x=[100223234]x

  17. x=[10018742795]x

  18. x=[100131241]x

  19. x=[14020100612160401]x

  20. x=[2101021000210002]x

  21. x=[1400130012100101]x

  22. x=[13700140013006141]x

In Problems 23 through 32 the eigenvalues of the coefficient matrix A are given. Find a general solution of the indicated system x=Ax. Especially in Problems 29 through 32, use of a computer algebra system (as in the application material for this section) may be useful.

  1. x=[3981636516721629]x;λ=1, 3, 3,

  2. x=[2850100153360153057]x;λ=2, 3, 3

  3. x=[2174161012]x;λ=2, 2, 2

  4. x=[511130321]x;λ=3, 3, 3

  5. x=[3553138810]x;λ=2, 2, 2

  6. x=[15743416111775]x;λ=2, 2, 2

  7. x=[11127461151136226]x;λ=1, 1, 2, 2

  8. x=[2121035301322120274525]x;λ=1, 1, 2, 2

  9. x=[351243022831910309279323]x;λ=1, 1, 1, 1

  10. x=[111266303000932463309514831383018]x;λ=2, 2, 3, 3, 3

  11. The characteristic equation of the coefficient matrix A of the system

    x=[3410430100340043]x

    is

    ϕ(λ)=(λ26λ+25)2=0.

    Therefore, A has the repeated complex conjugate pair 3±4i of eigenvalues. First show that the complex vectors

    v1=[1i00]Tandv2=[001i]T

    form a length 2 chain {v1,v2} associated with the eigenvalue λ=34i. Then calculate the real and imaginary parts of the complex-valued solutions

    v1eλtand(v1t+v2)eλt

    to find four independent real-valued solutions of x=Ax.

  12. The characteristic equation of the coefficient matrix A of the system

    x=[2083181009325933109032]x

    is

    ϕ(λ)=(λ24λ+13)2=0.

    Therefore, A has the repeated complex conjugate pair 2±3i of eigenvalues. First show that the complex vectors

    v1=[i3+3i01]T,v2=[310+9ii0]T

    form a length 2 chain {v1,v2} associated with the eigenvalue λ=2+3i. Then calculate (as in Problem 33) four independent real-valued solutions of x=Ax.

  13. Railway cars Find the position functions x1(t) and x2(t) of the railway cars of Fig. 7.5.2 if the physical parameters are given by

    m1=m2=c1=c2=c=k=1

    and the initial conditions are

    x1(0)=x2(0)=0,x1(0)=x2(0)=v0

    How far do the cars travel before stopping?

  14. Railway cars Repeat Problem 35 under the assumption that car 1 is shielded from air resistance by car 2, so now c1=0. Show that, before stopping, the cars travel twice as far as those of Problem 35.

  1. 37–46. Use the method of Examples 7 and 8 to find the Jordan normal form J of each coefficient matrix A given in Problems 23 through 32 (respectively).

7.6 Application Defective Eigenvalues and Generalized Eigenvectors

A typical computer algebra system can calculate both the eigenvalues of a given matrix A and the linearly independent (ordinary) eigenvectors associated with each eigenvalue. For instance, consider the 4×4 matrix

(1)A=[351243022831910309279323]

of Problem 31 in this section. When the matrix A has been entered, the Maple calculation

with(linalg): eigenvectors(A);
[1, 4, {[-1, 0, 1, 1], [0, 1, 3, 0]}]

or the Mathematica calculation

Eigensystem[A]
{{1,1,1,1},
{{-3,-1,0,3}, {0,1,3,0}, {0,0,0,0}, {0,0,0,0}}}

reveals that the matrix A in Eq. (1) has the single eigenvalue λ=1 of multiplicity 4 with only two independent associated eigenvectors v1 and v2. The Matlab command

[V, D] = eig(sym(A))

provides the same information. The eigenvalue λ=1 therefore has defect d=2. If B=A(1)I, you should find that B20 but B3=0. If

u1=[1000]T,u2=Bu1,u3=Bu2,

then {u1,u2,u3} should be a length 3 chain of generalized eigenvectors based on the ordinary eigenvector u3 (which should be a linear combination of the original eigenvectors v1 and v2). Use your computer algebra system to carry out this construction, and finally write four linearly independent solutions of the linear system x=Ax.

For a more exotic matrix to investigate, consider the Matlab’s gallery(5) example matrix

(2)A=[91121632527069141421168457557511493451138013891389177822334593365102410242048614424572].

Use appropriate commands like those illustrated here to show that A has a single eigenvalue λ=0 of multiplicity 5 and defect 4. Noting that A(0)I=A, you should find that A40 but that A5=0. Hence calculate the vectors

u1=[10000]T,u2=Au1,u3=Au2,u4=Au3,u5=Au4.

You should find that u5 is a nonzero vector such that Au5=0, and is therefore an (ordinary) eigenvector of A associated with the eigenvalue λ=0. Thus {u1,u2,u3,u4,u5} is a length 5 chain of generalized eigenvectors of the matrix A in Eq. (2), and you can finally write five linearly independent solutions of the linear system x=Ax.

7.7 Numerical Methods for Systems

We now discuss the numerical approximation of solutions of systems of differential equations. Our goal is to apply the methods of Sections 2.4 through 2.6 to the initial value problem

(1)x=f(t, x), x(t0)=x0

for a system of m first-order differential equations. In (1) the independent variable is the scalar t, and

x=(x1, x2, , xm)andf=(f1, f2, , fm)

are vector-valued functions. If the component functions of f and their first-order partial derivatives are all continuous in a neighborhood of the point (t0,x0), then Theorems 3 and 4 of Appendix A guarantee the existence and uniqueness of a solution x=x(t) of (1) on some subinterval [of the t-axis] containing t0. With this assurance we can proceed to discuss the numerical approximation of this solution.

Beginning with step size h, we want to approximate the values of x(t) at the points t1, t2, t3, , where tn+1=tn+h for n0. Suppose that we have already computed the approximations

x1,x2,x3,,xn

to the actual values

x(t1),x(t2),x(t3),,x(tn)

of the exact solution of the system in (1). We can then make the step from xn to the next approximation xn+1x(tn+1) by any one of the methods of Sections 2.4 through 2.6. Essentially all that is required is to write the iterative formula of the selected method in the vector notation of the present discussion.

Euler Methods for Systems

For example, the iterative formula of Euler’s method for systems is

(2)xn+1=xn+hf(t, xn).

To examine the case m=2 of a pair of first-order differential equations, let us write

x=[xy]andf=[fg].

Then the initial value problem in (1) is

(3)x=f(t,x,y),x(t0)=x0,y=g(t,x,y),y(t0)=y0,

and the scalar components of the vector formula in (2) are

(4)xn+1=xn+hf(tn,xn,yn),yn+1=yn+hg(tn,xm,yn).

Note that each iterative formula in (4) has the form of a single Euler iteration, but with yn inserted like a parameter in the first formula (for xn+1) and with xn inserted like a parameter in the second formula (for yn+1). The generalization to the system in (3) of each of the other methods in Sections 2.4 through 2.6 follows a similar pattern.

The improved Euler method for systems consists at each step of calculating first the predictor

(5)un+1=xn+hf(tn,xn)

and then the corrector

(6)xn+1=xn+h2[f(tn,xn)+f(tn+1,un+1)].

For the case of the two-dimensional initial value problem in (3), the scalar components of the formulas in (5) and (6) are

(7)un+1=xn+hf(tn,xn,yn),vn+1=yn+hg(tn,xn,yn)

and

(8)xn+1=xn+h2[f(tn,xn,yn)+f(tn+1,un+1,vn+1)],yn+1=yn+h2[g(tn,xn,yn)+g(tn+1,un+1,vn+1)].

Example 1

Consider the initial value problem

(9)x=3x2y,x(0)=3;y=5x4y,y(0)=6.

The exact solution of the system in (9) is

(10)x(t)=2e2t+et,y(t)=5e2t+et.

Here we have f(x, y)=3x2y and g(x, y)=5x4y in (3), so the Euler iterative formulas in (4) are

xn+1=xn+h(3xn2yn),yn+1=yn+h(5xn4yn).

With step size h=0.1 we calculate

x1=3+(0.1)[3.32.6]=2.7,y1=6+(0.1)[5.34.6]=5.1

and

x2=2.7+(0.1)[3(2.7)2(5.1)]=2.49y2=5.1+(0.1)[5(2.7)4(5.1)]=4.41.

The actual values at t2=0.2 given by (10) are x(0.2)2.562 and y(0.2)4.573.

To compute the improved Euler approximations to x(0.2) and y(0.2) with a single step of size h=0.2, we first calculate the predictors

u1=3+(0.2)[3.32.6]=2.4,v1=6+(0.2)[5.34.6]=4.2.

Then the corrector formulas in (8) yield

x1=3+(0.1)([3.32.6]+[3(2.4)2(4.2)])=2.58,y1=6+(0.1)([5.34.6]+[5(2.4)4(4.2)])=4.62.

As we would expect, a single improved Euler step gives better accuracy than two ordinary Euler steps.

The Runge–Kutta Method and Second-Order Equations

The vector version of the iterative formula for the Runge–Kutta method is

(11)xn+1=xn+h6(k1+2k2+2k3+k4),

where the vectors k1, k2, k3, and k4 are defined (by analogy with Eqs. (5a)(5d) of Section 2.6) as follows:

(12)k1=f(tn,xn),k2=f(tn+12h,xn+12hk1),k3=f(tn+12h,xn+12hk2),k4=f(tn+h,xn+hk3).

To describe in scalar notation the Runge–Kutta method for the two-dimensional initial value problem

(3)x=f(t,x,y),x(t0)=x0,y=g(t,x,y),y(t0)=y0,

let us write

x=[xy],f=[fg],andki=[FiGi].

Then the Runge–Kutta iterative formulas for the step from (xn,yn) to the next approximation (xn+1,yn+1)(x(tn+1), y(tn+1)) are

(13)xn+1=xn+h6(F1+2F2+2F3+F4),yn+1=yn+h6(G1+2G2+2G3+G4),

where the values F1, F2, F3, and F4 of the function f are

(14)F1=f(tn,xn,yn),F2=f(tn+12h,xn+12hF1,yn+12hG1),F3=f(tn+12h,xn+12hF2,yn+12hG2),F4=f(tn+h,xn+hF3,yn+hG3);

G1, G2, G3, and G4 are the similarly defined values of the function g.

Perhaps the most common application of the two-dimensional Runge–Kutta method is to the numerical solution of second-order initial value problems of the form

(15)x=g(t,x,x),x(t0)=x0,x(t0)=y0.

If we introduce the auxiliary variable y=x, then the problem in (15) translates into the two-dimensional first-order problem

(16)x=y,x(t0)=x0,y=g(t,x,y),y(t0)=y0.

This is a problem of the form in (3) with f(t,x,y)=y.

If the functions f and g are not too complicated, then it is feasible to carry out manually a reasonable number of steps of the two-dimensional Runge–Kutta method described here. But the first operating electronic computers were constructed (during World War II) specifcally to implement methods similar to the Runge–Kutta method for the numerical computation of trajectories of artillery projectiles. The application material for this section lists TI-Nspire CX CAS and Python versions of Program RK2DIM that can be used with two-dimensional systems.

Example 2

The exact solution of the initial value problem

(17)x=x;x(0)=0,x(0)=1

is x(t)=sint. The substitution y=x translates (17) into the two-dimensional problem

(18)x=y,x(0)=0;y=x,y(0)=1,

which has the form in (3) with f(t,x,y)=y and g(t,x,y)=x. The table in Fig.7.7.1 shows the results produced for 0t5 (radians) using Program RK2DIM with step size h=0.05. The values shown for x=sint and y=cost are all accurate to five decimal places.

FIGURE 7.7.1.

Runge–Kutta values (with h=0.05) for the problem in Eq. (18).

t x=sint y=cost
0.5 +0.47943 +0.87758
1.0 +0.84147 +0.54030
1.5 +0.99749 +0.07074
2.0 +0.90930 0.41615
2.5 +0.59847 0.80114
3.0 +0.14112 0.98999
3.5 0.35078 0.93646
4.0 0.75680 0.65364
4.5 0.97753 0.21080
5.0 0.95892 +0.28366

Example 3

Lunar lander In Example 4 of Section 2.3 we considered a lunar lander that initially is falling freely toward the surface of the moon. Its retrorockets, when fired, provide a deceleration of T=4m/s2. We found that a soft touchdown at the lunar surface is achieved by igniting these retrorockets when the lander is at a height of 41,870 meters (just over 26 miles) above the surface and is then descending at the rate of 450 m/s.

Now we want to compute the descent time of the lunar lander. Let the distance x(t) of the lander from the center of the moon be measured in meters and measure time t in seconds. According to the analysis in Section 2.3 (where we used r(t) in stead of x(t)), x(t) satisfes the initial value problem

(19)d2xdt2=TGMx2=44.9044×1012x2,x(0)=R+41870=1,781,870,x(0)=450

where G6.6726×1011 N(m/kg)2 is the universal gravitational constant and M=7.35×1022 kg and R=1.74×106 m are the mass and radius of the moon. We seek the value of t when x(t)=R=1,740,000..

The problem in (19) is equivalent to the first-order system

(20)dxdt=y,x(0)=1,781,870;dydt=44.9044×10x2,y(0)=450.

The table in Fig. 7.7.2 shows the result of a Runge–Kutta approximation with step size h=1 (the indicated data agreeing with those obtained with step size h=2). Evidently, touchdown on the lunar surface (x=1,740,000) occurs at some time between t=180 and t=190 seconds. The table in Fig. 7.7.3 shows a second Runge–Kutta approximation with t(0)=180, x(0)=1,740,059, y(0)=16.83, and h=0.1. Now it is apparent that the lander’s time of descent to the lunar surface is very close to 187 seconds; that is, 3 min 7 s. (The final velocity terms in these two tables are positive because the lander would begin to ascend if its retrorockets were not turned off at touchdown.)

FIGURE 7.7.2.

The lander’s descent to the lunar surface.

t(s) x(m) v(m/s)
0 1,781,870 450.00
20 1,773,360 401.04
40 1,765,826 352.37
60 1,759,264 303.95
80 1,753,667 255.74
100 1,749,033 207.73
120 1,745,357 159.86
140 1,742,637 112.11
160 1,740,872 64.45
180 1,740,059 16.83
200 1,740,199 30.77

FIGURE 7.7.3.

Focusing on the lunar lander’s soft touchdown.

t(s) x(m) v(m/s)
180 1,740,059 16.83
181 1,740,044 14.45
182 1,740,030 12.07
183 1,740,019 9.69
184 1,740,011 7.31
185 1,740,005 4.93
186 1,740,001 2.55
187 1,740,000 0.17
188 1,740,001 2.21
189 1,740,004 4.59
190 1,740,010 6.97

Higher-Order Systems

As we saw in Section 7.1, any system of higher-order differential equations can be replaced with an equivalent system of first-order differential equations. For example, consider the system

(21)x=F(t,x,y,x,y),y=G(t,x,y,x,y)

of second-order equations. If we substitute

x=x1,y=x2,x=x3=x1,y=x4=x2,

then we get the equivalent system

(22)x1=x3,x2=x4,x3=F(t,x1,x2,x3,x4),x4=G(t,x1,x2,,x3,x4)

of four first-order equations in the unknown functions x1(t)=x(t), x2(t)=y(t), x3(t), and x4(t). It would be a routine (if slightly tedious) matter to write a four-dimensional version of program RK2DIM for the purpose of solving such a system. But in a programming language that accommodates vectors, an n-dimensional Runge–Kutta program is scarcely more complicated than a one-dimensional program. For instance, the application material for this section lists the n-dimensional Matlab program rkn that closely resembles the one-dimensional program rk of Fig.2.6.11.

Example 4

Batted baseball Suppose that a batted ball starts at x0=0, y0=0 with initial velocity v0=160 ft/s and with initial angle of inclination θ=30. If air resistance is ignored, we find by the elementary methods of Section 1.2 that the baseball travels a [horizontal] distance of 4003 ft (approximately 693 ft) in 5 s before striking the ground. Now suppose that in addition to a downward gravitational acceleration (g=32 ft/s2), the baseball experiences an acceleration due to air resistance of (0.0025)v2 feet per second per second, directed opposite to its instantaneous direction of motion. Determine how far the baseball will travel horizontally under these conditions.

Solution

According to Problem 30 of Section 7.1, the equations of motion of the baseball are

(23)d2xdt2=cvdxdt,d2ydt2=cvdydtg

where v=(x)2+(y)2 is the speed of the ball, and where c=0.0025 and g=32 in fps units. We convert to a first-order system as in (22) and thereby obtain the system

(24)x1=x3,x2=x4,x3=cx3x32+x42,x4=cx4x32+x42g

of four first-order differential equations with

(25)x1(0)=x2(0)=0,x3(0)=803,x4(0)=80.

Note that x3(t) and x4(t) are simply the x- and y-components of the baseball’s velocity vector, so v=x32+x42. We proceed to apply the Runge–Kutta method to investigate the motion of the batted baseball described by the initial value problem in (24) and (25), first taking c=0 to ignore air resistance and then using c=0.0025 to take air resistance into account.

Without Air Resistance: Figure 7.7.4 shows the numerical results obtained when a Runge–Kutta program such as rkn is applied with step size h=0.1 and with c=0 (no air resistance). For convenience in interpreting the results, the printed output at each selected step consists of the horizontal and vertical coordinates x and y of the baseball, its velocity v, and the angle of inclination α of its velocity vector (in degrees measured from the horizontal). These results agree with the exact solution when c=0. The ball travels a horizontal distance of 4003692.82 ft in exactly 5 s, having reached a maximum height of 100 ft after 2.5 s. Note also that the ball strikes the ground at the same angle and with the same speed as its initial angle and speed.

FIGURE 7.7.4.

The batted baseball with no air resistance (c=0).

t x y v α
0.0 0.00 0.00 160.00 +30
0.5 69.28 36.00 152.63 +25
1.0 138.56 64.00 146.64 +19
1.5 207.85 84.00 142.21 +13
2.0 277.13 96.00 139.48 +7
2.5 346.41 100.00 138.56 +0
3.0 415.69 96.00 139.48 7
3.5 484.97 84.00 142.21 13
4.0 554.26 64.00 146.64 19
4.5 623.54 36.00 152.63 25
5.0 692.82 0.00 160.00 30

FIGURE 7.7.5.

The batted baseball with air resistance (c=0.0025).

t x y v α
0.0 0.00 0.00 160.00 +30
0.5 63.25 32.74 127.18 +24
1.0 117.11 53.20 104.86 +17
1.5 164.32 63.60 89.72 +8
2.0 206.48 65.30 80.17 3
2.5 244.61 59.22 75.22 15
3.0 279.29 46.05 73.99 27
3.5 310.91 26.41 75.47 37
4.0 339.67 0.91 78.66 46

FIGURE 7.7.6.

The batted ball’s apex and its impact with the ground.

t x y v α  
1.5 164.32 63.60 89.72 +8  
1.6 173.11 64.60 87.40 +5  
1.7 181.72 65.26 85.29 +3  
1.8 190.15 65.60 83.39 +1 ← Apex
1.9 198.40 65.61 81.68 1
2.0 206.48 65.30 80.17 3  
 
3.8 328.50 11.77 77.24 42  
3.9 334.14 6.45 77.93 44  
4.0 339.67 0.91 78.66 46 ← Impact
4.1 345.10 4.84 79.43 47
4.2 350.41 10.79 80.22 49  

With Air Resistance: Figure 7.7.5 shows the results obtained with the fairly realistic value of c=0.0025 for the air resistance for a batted baseball. To within a hundredth of a foot in either direction, the same results are obtained with step sizes h=0.05 and h=0.025. We now see that with air resistance the ball travels a distance well under 400 ft in just over 4 s. The more refined data in Fig.7.7.6 show that the ball travels horizontally only about 340 ft and that its maximum height is only about 66 ft. As illustrated in Fig.7.7.7, air resistance has converted a massive home run into a routine fly ball (if hit straightaway to center field). Note also that when the ball strikes the ground, it has slightly under half its initial speed (only about 79 ft/s) and is falling at a steeper angle (about 46). Every baseball fan has observed empirically these aspects of the trajectory of a fly ball.

FIGURE 7.7.7.

An “easy out” or a home run?

Variable Step Size Methods

The Runge–Kutta method for a large system requires an appreciable amount of computational labor, even when a computer is employed. Therefore, just as the step size h should not be so large that the resulting error in the solution is unacceptable, h ought not to be so small that too many steps are needed, hence requiring an unacceptable amount of computation. Thus the practical numerical solution of differential equations involves a tradeoff between accuracy and efficiency.

To facilitate this tradeoff, modern variable step size methods vary the step size h as the solution process proceeds. Large steps are taken in regions where the dependent variables are changing slowly; smaller steps are taken when these variables are changing rapidly, in order to prevent large errors.

An adaptable or variable step size Runge–Kutta method employs both a pre-assigned minimum error tolerance MinTol and a maximum error tolerance MaxTol to attempt to ensure that the error made in the typical step from xnto xn+1 is neither too large (and hence inaccurate) nor too small (and hence in efficient). A fairly simple scheme for doing this may be outlined as follows:

The detailed implementation of such a scheme can be complicated. For a much more complete but readable discussion of adaptive Runge–Kutta methods, see Section 17.2 of William H. Press et al., Numerical Recipes: The Art of Scientific Computing (Cambridge: Cambridge: University Press, 2007).

Several widely available scientific computing packages (such as Maple, Mathematica, and Matlab) include sophisticated variable step size programs that will accommodate an essentially arbitrary number of simultaneous differential equations. Such a general-purpose program might be used, for example, to model numerically the major components of the solar system:the sun and the nine (known) major planets. If mi denotes the mass and ri=(xi, yi, zi) denotes the position vector of the ith one of these 10 bodies, then—by Newton’s laws—the equation of motion of mi is

(26)mir=jiGmimj(rij)3(rjri),

where rij=|rjri| denotes the distance between mi and mj. For each i=1, 2, , 10, the summation in Eq. (26) is over all values of ji from 1 to 10. The 10 vector equations in (26) constitute a system of 30 second-order scalar equations, and the equivalent first-order system consists of 60 differential equations in the coordinates and velocity components of the 10 major bodies in the solar system. Mathematical models that involve this many (or more) differential equations—and that require sophisticated software and hardware for their numerical analysis—are quite common in science, engineering, and applied technology.

Earth–Moon Satellite Orbits

For an example of a program whose efficient solution requires adaptive step size methods, we consider an Apollo satellite in orbit about the Earth E and Moon M. Figure 7.7.8 shows an x1x2-coordinate system whose origin lies at the center of mass of the Earth and the Moon and which rotates at the rate of one revolution per “moon month”of approximately τ=27.32 days, so the Earth and Moon remain fixed in their positions on the x1-axis. If we take as unit distance the distance (about 384,000 kilometers, assumed constant) between the Earth and Moon centers, then their coordinates are E(μ, 0) and M(1μ,0), where μ=mM/(mE+mM) in terms of the Earth mass mE and Moon mass mM. If we take the total mass mE+mM as the unit of mass and τ/(2π)4.35 days as the unit of time, then the gravitational constant is G=1 in Eq. (26), and the equations of motion of the satellite position S(x1, x2) are

(27)x1=x1+2x2(1μ)(x1+μ)(rE)3μ(x11+μ)(rM)3,x2=x22x1(1μ)x2(rE)3μx2(rM)3,

FIGURE 7.7.8.

The Earth–Moon center-of-mass coordinate system.

where rE and rM denote the satellite’s distance to the Earth and Moon (indicated in Fig. 7.7.8). The initial two terms on the right-hand side of each equation result from the rotation of the coordinate system. In the system of units described here, the lunar mass is approximately mM=0:012277471. The second-order system in (27) can be converted to an equivalent first-order system (of four differential equations) by substituting

x1=x3,x2=x4,so thatx1=x3,  x2=x4.

Suppose that the satellite initially is in a clockwise circular orbit of radius about 2400 kilometers about the Moon. At its farthest point from the Earth (x1=0.994) it is “launched” into Earth–Moon orbit with initial velocity v0. The corresponding initial conditions are

x1(0)=0.994,x2(0)=0,x3(0)=0,x4(0)=v0.

An adaptive step size method (ode45) in the Matlab software system was used to solve numerically the system in (27). The orbits in Figs. 7.7.9 and 7.7.10 were obtained with

v0=2.031732629557andv0=2.001585106379,

FIGURE 7.7.9.

Apollo Moon–Earth bus orbit with insertion velocity v0=7476 km/h.

FIGURE 7.7.10.

Apollo Moon–Earth bus orbit with insertion velocity v0=7365 km/h.

respectively. [In the system of units used here, the unit of velocity is approximately 3680 km/h.] In each case a closed but multilooped periodic trajectory about the Earth and the Moon—a so-called bus orbit—is obtained, but a relatively small change in the initial velocity changes the number of loops! For more information, see NASA Contractor Report CR-61139, “Study of the Methods for the Numerical Solution of Ordinary Differential Equations,” prepared by O. B. Francis, Jr. et al. for the NASA–George C. Marshall Space Flight Center, June 7, 1966.

So-called Moon–Earth “bus orbits” are periodic—that is, are closed trajectories traversed repeatedly by the satellite—only in a rotating x1x2-coordinate system as discussed above. The satellite of Fig. 7.7.9 traverses its closed orbit and returns to rendezvous with the Moon about 48.4 days after its insertion into orbit. Figures 7.7.11 and 7.7.12 illustrate the motion of the same satellite—but in an ordinary nonrotating xy-coordinate system centered at the Earth, in which the Moon encircles the Earth counterclockwise in a near-circular orbit, completing one revolution in about 27.3 days. The Moon starts at point S, and after 48.4 days it has completed a bit over 1.75 revolutions about the Earth and reaches the point R at which its rendezvous with the satellite occurs. Figure 7.7.11 shows the positions of Moon and satellite a day and a half after the satellite’s insertion into its orbit, each traveling in a generally counterclockwise direction around the Earth. Figure 7.7.12 shows their positions a day and a half before their rendezvous at point R, the satellite meanwhile having encircled the Earth about 2.5 times in an orbit that (in the indicated xy-coordinate system) appears to resemble a slowly varying ellipse.

FIGURE 7.7.11.

The moon and satellite in a nonrotating coordinate system, 1.5 days after orbital insertion of the satellite at starting point S.

FIGURE 7.7.12.

The moon and satellite in a nonrotating coordinate system, 1.5 days before their rendezvous at point R.

7.7 Problems

A hand-held calculator will suffice for Problems 1 through 8. In each problem an initial value problem and its exact solution are given. Approximate the values of x(0.2) and y(0.2) in three ways: (a) by the Euler method with two steps of size h=0.1; (b) by the improved Euler method with a single step of size h=0.2; and (c) by the Runge–Kutta method with a single step of size h=0.2. Compare the approximate values with the actual values x(0.2) and y(0.2).

  1. x=x+2y, x(0)=0,y=2x+y, y(0)=2;x(t)=e3tet, y(t)=e3t+et

  2. x=2x+3y,x(0)=1,y=2x+y, y(0)=1;x(t)=et, y(t)=et

  3. x=3x+4y, x(0)=1,y=3x+2y,y(0)=1;x(t)=17(8e6tet), y(t)=17(6e6t+et)

  4. x=9x+5y, x(0)=1,y=6x2y, y(0)=0;x(t)=5e3t+6e4t, y(t)=6e3t6e4t

  5. x=2x5y, x(0)=2,y=4x2y, y(0)=3;x(t)=2cos4t114 sin 4t, y(t)=3 cos 4t+12 sin 4t

  6. x=x2y, x(0)=0,y=2x+y, y(0)=4;x(t)=4et sin 2t, y(t)=4et cos 2t

  7. x=3xy, x(0)=2,y=x+y, y(0)=1;x(t)=(t+2)e2t, y(t)=(t+1)e2t

  8. x=5x9y, x(0)=0,y=2xy, y(0)=1;x(t)=3e2t sin 3t, y(t)=e2t(sin 3tcos 3t)

A computer will be required for the remaining problems in this section. In Problems 9 through 12, an initial value problem and its exact solution are given. In each of these four problems, use the Runge–Kutta method with step sizes h=0.1 and h=0.05 to approximate to five decimal places the values x(1) and y(1) Compare the approximations with the actual values.

  1. x=2xy,x(0)=1,y=x+2y,y(0)=0;x(t)=e2t cos t,y(t)=e2t sin t

  2. x=x+2y,x(0)=0,y=x+et,y(0)=0;x(t)=19(2e2t2et+6tet),y(t)=19(e2tet+6tet)

  3. x=x+y(1+t3)et,x(0)=0,y=xy(t3t2)et,y(0)=1;x(t)=et(sin tt),y(t)=et(cos t+t3)

  4. x+x=sin t,x(0)=0;x(t)=12(sin tt cos t)

  5. Crossbow bolt Suppose that a crossbow bolt is shot straight upward with initial velocity 288 ft/s. If its deceleration due to air resistance is (0.04)v, then its height x(t) satisfies the initial value problem

    x=32(0.04)x;x(0)=0,x(0)=288.

    Find the maximum height that the bolt attains and the time required for it to reach this height.

  6. Repeat Problem 13, but assume instead that the deceleration of the bolt due to air resistance is (0.0002)v2.

  7. Height of projectile Suppose that a projectile is fired straight upward with initial velocity v0 from the surface of the earth. If air resistance is not a factor, then its height x(t) at time t satisfies the initial value problem

    d2xdt2=gR2(x+R)2;x(0)=0,x(0)=v0.

    Use the values g=32.15 ft/s2  0.006089 mi/s2 for the gravitational acceleration of the earth at its surface and R=3960 mi as the radius of the earth. If v0=1 mi/s, find the maximum height attained by the projectile and its time of ascent to this height.

Batted Baseball

Problems 16 through 18 deal with the batted baseball of Example 4, having initial velocity 160 ft/s and air resistance coefficient c=0.0025.

  1. Find the range—the horizontal distance the ball travels before it hits the ground—and its total time of flight with initial inclination angles 40, 45, and 50.

  2. Find (to the nearest degree) the initial inclination that maximizes the range. If there were no air resistance it would be exactly 45, but your answer should be less than 45.

  3. Find (to the nearest half degree) the initial inclination angle greater than 45 for which the range is 300 ft.

  4. Home run Home run Find the initial velocity of a baseball hit by Babe Ruth (with c=0.0025 and initial inclination 40) if it hit the bleachers at a point 50 ft high and 500 horizontal feet from home plate.

  5. Crossbow bolt Crossbow bolt Consider the crossbow bolt of Problem 14, fired with the same initial velocity of 288 ft/s and with the air resistance deceleration (0.0002)v2 directed opposite its direction of motion. Suppose that this bolt is fired from ground level at an initial angle of 45. Find how high vertically and how far horizontally it goes, and how long it remains in the air.

  6. Artillery projectile Artillery projectile Suppose that an artillery projectile is fired from ground level with initial velocity 3000 ft/s and initial inclination angle 40. Assume that its air resistance deceleration is (0.0001)v2. (a) What is the range of the projectile and what is its total time of flight? What is its speed at impact with the ground? (b) What is the maximum altitude of the projectile, and when is that altitude attained? (c) You will find that the projectile is still losing speed at the apex of its trajectory. What is the minimum speed that it attains during its descent?

7.7 Application Comets and Spacecraft

Figure 7.7.13 lists TI-Nspire CX CAS and Python versions of the two-dimensional Runge–Kutta program RK2DIM. You should note that it closely parallels the one-dimensional Runge–Kutta program listed in Fig. 2.6.11, with a single line there replaced (where appropriate) with two lines here to calculate a pair of x- and y-values or slopes. Note also that the notation used is essentially that of Eqs. (13) and (14) in this section. The first several lines define the functions and initial data needed for Example 1.

Figure 7.7.14 exhibits an n-dimensional Matlab implementation of the Runge–Kutta method. The Matlab function f defines the vector of right-hand sides of the differential equations in the system x=f(t,x) to be solved. The rkn function then takes as input the initial t-value t, the column vector x of initial x-values, the final t-value t1, and the desired number n of subintervals. As output it produces the resulting column vector T of t-values and the matrix X whose rows give the corresponding x-values. For instance, with f as indicated in the figure, the Matlab command

[T,X] = rkn(0, [0;1], 5, 50)

then generates the data shown in the table of Fig. 7.7.1 (which lists only every fifth value of each variable).

You can use Examples 1 through 3 in this section to test your own implementation of the Runge–Kutta method. Then investigate the comet and spacecraft problems described next. Additional application material at the Expanded Applications site indicated in the margin describes additional numerical ODE investigations ranging from batted baseballs to the Apollo orbits shown in Figs. 7.7.9 and 7.7.10.

Your Spacecraft Landing

Your spacecraft is traveling at constant velocity V, approaching a distant earthlike planet with mass M and radius R. When activated, your deceleration system provides a constant thrust T until impact with the surface of the planet. During the period of deceleration, your distance x(t) from the center of the planet satisfies the

FIGURE 7.7.13.

TI-Nspire CX CAS and Python two-dimensional Runge–Kutta programs.

TI-Nspire CX CAS Python Comment
Define rk2dim()=Prgm
   f(t,x,y):=y
   g(t,x,y):=-x
   n:=50
   t:=0.0
   x:=0.0
   y:=1.0
   t1:=5.0
   h:=(t1-t)/n
   For i,1,n
        t0:=t
        x0:=x
        y0:=y
        f1:=f(t,x,y)
        g1:=g(t,x,y)
        t:=t0+h/2
        x:=x0+(h*f1)/2
        y:=y0+(h*g1)/2
        f2:=f(t,x,y)
        g2:=g(t,x,y)
        x:=x0+(h*f2)/2
        y:=y0+(h*g2)/2
        f3:=f(t,x,y)
        g3:=g(t,x,y)
        t:=t0+h
        x:=x0+h*f3
        y:=y0+h*g3
        f4:=f(t,x,y)
        g4:=g(t,x,y)
        fa:=(f1+2*f2+2*f3+f4)/6
        ga:=(g1+2*g2+2*g3+g4)/6
        x:=x0+h*fa
        y:=y0+h*ga
        Disp t,x,y
   EndFor
EndPrgm
# Program RK2DIM
def F(T,X,Y): return Y
def G(T,X,Y): return -X
N = 50
T = 0.0
X = 0.0
Y = 1.0
T1 = 5
H = (T1-T)/N
for I in range(N):
T0 = T
X0 = X
Y0 = Y
F1 = F(T,X,Y)
G1 = G(T,X,Y)
T = T0 + H/2
X = X0 + H*F1/2
Y = Y0 + H*G1/2
F2 = F(T,X,Y)
G2 = G(T,X,Y)
X = X0 + H*F2/2
Y = Y0 + H*G2/2
F3 = F(T,X,Y)
G3 = G(T,X,Y)
T = T0 + H
X = X0 + H*F3
Y = Y0 + H*G3
F4 = F(T,X,Y)
G4 = G(T,X,Y)
FA = (F1+2*F2+2*F3+F4)/6
GA = (G1+2*G2+2*G3+G4)/6
X = Y0 + H*FA
Y = Y0 + H*GA
print (T,X,Y)
# END
Program title
Define function f
Define function g
No. of steps
Initial t
Initial x
Initial y
Final t
Step size
Begin loop
Save previous t
Save previous x
Save previous y
First f-slope
First g-slope
Midpoint t
Midpt x-predictor
Midpt y-predictor
Second f-slope
Second g-slope
Midpt x-predictor
Midpt y-predictor
Third f-slope
Third g-slope
New t
Endpt x-predictor
Endpt y-predictor
Fourth f-slope
Fourth g-slope
Average f-slope
Average g-slope
x-corrector
y-corrector
Display results
End loop
function xp = f(t,x)
xp = x;
xp(1) = x(2);
xp(2) = -x(1);
function [T,Y] = rkn(t,x,t1,n)
h = (t1 - t)/n;                   % step size
T = t;                            % initial t
X = x′;                           % initial x-vector
for i = 1:n                       % begin loop
   k1 = f(t,x);                   % first k-vector
   k2 = f(t+h/2,x+h*k1/2);        % second k-vector
   k3 = f(t+h/2,x+h*k2/2);        % third k-vector
   k4 = f(t+h ,x+h*k3);           % fourth k-vector
   k = (k1+2*k2+2*k3+k4)/6;       % average k-vector
   t = t + h;                     % new t
   x = x + h*k;                   % new x
   T = [T;t];                     % update t-column
   X = [X;x′];                    % update x-matrix
   end                            % end loop

FIGURE 7.7.14.

Matlab implementation of the Runge–Kutta method.

differential equation

(1)d2xdt2=TGx2,

where G6.6726×1011N(m/kg)2 as in Example 3. Your question is this: At what altitude above the surface should your deceleration systembe activated in order to achieve a soft touchdown? For a reasonable problem, you can take

M=5.97×1024(kg),R=6.38×106(m),V=p×104(km/h),T=g+q(m/s2)

where g=GM/R2 is the surface gravitational acceleration of the planet. Choose p to be the smallest nonzero digit and q the next-to-smallest nonzero digit in your ID number. Find the “ignition altitude” accurate to the nearest meter and the resulting “descent time” accurate to the nearest tenth of a second.

Kepler’s Law of Planetary (or Satellite) Motion

Consider a satellite in elliptical orbit around a planet of mass M, and suppose that physical units are so chosen that GM=1 (where G is the gravitational constant). If the planet is located at the origin in the xy-plane, then the equations of motion of the satellite are

(2)d2xdt2=x(x2+y2)3/2, d2ydt2=y(x2+y2)3/2.

Let T denote the period of revolution of the satellite. Kepler’s third law says that the square of T is proportional to the cube of the major semiaxis a of its elliptical orbit. In particular, if GM=1, then

(3)T2=4π2a3.

(For details, see Section 11.6 of Edwards and Penney, Calculus: Early Transcendentals, 7th ed., Hoboken, NJ: Pearson, 2008).) If the satellite’s x- and y-components of velocity, x3=x=x1 and x4=y=x2, are introduced, then the system in (2) translates into a system of four first-order differential equations having the form of those in Eq. (22) of this section.

  1. Solve this 4×4 system numerically with the initial conditions

    x(0)=1,y(0)=0,x(0)=0,y(0)=1

    that correspond theoretically to a circular orbit of radius a=1, so Eq. (3) gives T=2π. Is this what you get?

  2. Now solve the system numerically with the initial conditions

    x(0)=1,y(0)=0,x(0)=0,y(0)=126

    that correspond theoretically to an elliptical orbit with major semiaxis a a=2, so Eq. (3) gives T=4π2. Is this what you get?

Halley’s Comet

Halley’s comet last reached perihelion (its point of closest approach to the sun at the origin) on February 9, 1986. Its position and velocity components at that time were

p0=(0.325514,0.459460,0.166229)andv0=(9.096111,6.916686,1.305721)

(respectively), with position in AU (astronomical units, in which the unit of distance is the major semiaxis of the earth’s orbit) and time in years. In this system, the threedimensional equations of motion of the comet are

(4)d2xdt2=μxr3, d2ydt2=μyr3, d2zdt2=μzr3

where

μ=4π2andr=x2+y2+z2.

Solve the equations in (4) numerically to verify the appearance of the yz-projection of the orbit of Halley’s comet shown in Fig. 7.7.15. Plot the xy- and xz-projections as well.

FIGURE 7.7.15.

yz-projection of the orbit of Halley’s comet.

Figure 7.7.16 shows the graph of the distance r(t) of Halley’s comet from the sun. Inspection of this graph indicates that Halley’s comet reaches a maximum distance (at aphelion) of about 35 AU in a bit less than 40 years and returns to perihelion after about three-quarters of a century. The closer look in Fig. 7.7.17 indicates that the period of revolution of Halley’s comet is about 76 years. Use your numerical solution to refine these observations. What is your best estimate of the calendar date of the comet’s next perihelion passage?

FIGURE 7.7.16.

200-year plot of the distance r(t) of Halley’s comet from the sun. Is there a cusp near t=75?

FIGURE 7.7.17.

A closer look at Halley’s perihelion passage after about 76 years.

Your Own Comet

The night before your birthday in 2007 you set up your telescope on a nearby mountaintop. It was a clear night, and you had a stroke of luck: At 12:30 A.M. you spotted a new comet. After repeating the observation on successive nights, you were able to calculate its solar system coordinates p0=(x0, y0, z0) and its velocity vector v0=(vx0, vy0, vz0) on that first night. Using this information, determine the following:

Using units of length in AU and time in earth years, the equations of motion of your comet are given in (4). For your personal comet, begin with random initial position and velocity vectors with the same order of magnitude as those of Halley’s comet. Repeat the random selection of initial position and velocity vectors, if necessary, until you get a plausible eccentric orbit that ranges well outside the earth’s orbit (as most real comets do).

8 Matrix Exponential Methods

8.1 Matrix Exponentials and Linear Systems

The solution vectors of an n×n homogeneous linear system

(1)x=Ax

can be used to construct a square matrix X=Φ(t) that satisfies the matrix differential equation

(1′)X=AX

associated with Eq. (1). Suppose that x1(t), x2(t),, xn(t) are n linearly independent solutions of Eq. (1). Then the n×n matrix

(2)Φ(t)=[|||x1(t)x2(t)xn(t)|||]

having these solution vectors as its column vectors, is called a fundamental matrix for the system in (1).

Fundamental Matrix Solutions

Because the column vector x=xj(t) of the fundamental matrix Φ(t) in (2) satisfies the differential equation x=Ax, it follows (from the definition of matrix multiplication) that the matrix X=Φ(t) itself satisfies the matrix differential equation X=AX. Because its column vectors are linearly independent, it also follows that the fundamental matrix Φ(t) is nonsingular, and therefore has an inverse matrix Φ(t)1. Conversely, any nonsingular matrix solution Ψ(t) of Eq. (1′) has linearly independent column vectors that satisfy Eq. (1), so Ψ(t) is a fundamental matrix for the system in (1).

In terms of the fundamental matrix Φ(t) in (2), the general solution

(3)x(t)=c1x1(t)+c2x2(t)++cnxn(t)

of the system x=Ax can be written in the form

(4)x(t)=Φ(t)c

where c=[c1c2cn]T is an arbitrary constant vector. If Ψ(t) is any other fundamental matrix for (1), then each column vector of Ψ(t) is a linear combination of the column vectors of Φ(t), so it follows from Eq. (4) that

(4′)Ψ(t)=Φ(t)C

for some n×n matrix C of constants.

In order that the solution x(t) in (3) satisfy a given initial condition

(5)x(0)=x0,

it suffices that the coefficient vector c in (4) be such that Φ(0)c=x0; that is, that

(6)c=Φ(0)1x0.

When we substitute (6) in Eq. (4), we get the conclusion of the following theorem.

Section 7.3 tells us how to find a fundamental matrix for the system

(9)x=Ax

with constant n×n coefficient matrix A, at least in the case where A has a complete set of n linearly independent eigenvectors v1, v2, , vn associated with the (not necessarily distinct) eigenvalues λ1, λ2, ,λn, respectively. In this event the corresponding solution vectors of Eq. (9) are given by

xi(t)=vieλit

for i=1, 2, , n. Therefore, the n×n matrix

(10)Φ(t)=[|||v1eλ1tv2eλ2tvneλnt|||]

having the solutions x1, x2, , xn as column vectors is a fundamental matrix for the system x=Ax.

In order to apply Eq. (8), we must be able to compute the inverse matrix Φ(0)1. The inverse of the nonsingular 2×2 matrix

A=[abcd]

is

(11)A1=1Δ[dbca],

where Δ=det(A)=adbc0. The inverse of the nonsingular 3×3 matrix A=[aij] is given by

(12)A1=1Δ[+A11A12+A13A21+A22A23+A31A32+A33]T,

where Δ=det(A)0 and Aij denotes the determinant of the 2×2 submatrix of A obtained by deleting the i th row and j th column of A. (Do not overlook the symbol T for transpose in Eq. (12).) The formula in (12) is also valid upon generalization to n×n matrices, but in practice inverses of larger matrices are usually computed instead by row reduction methods (see any linear algebra text) or by using a calculator or computer algebra system.

Example 1

Find a fundamental matrix for the system

(13)x=4x+2y,y=3xy,

and then use it to find the solution of (13) that satisfies the initial conditions x(0)=1, y(0)=1.

Solution

The linearly independent solutions

x1(t)=[e2t3e2t]andx2(t)=[2e5te5t]

found in Example 1 of Section 7.3 yield the fundamental matrix

(14)Φ(t)=[e2t2e5t3e2te5t].

Then

Φ(0)=[1231],

and the formula in (11) gives the inverse matrix

(15)Φ(0)1=17[1231].

Hence the formula in (8) gives the solution

x(t)=[e2t2e5t3e2te5t](17)[1231][11]=(17)[e2t2e5t3e2te5t][32],

and so

x(t)=17[3e2t+4e5t9e2t+2e5t].

Thus the solution of the original initial value problem is given by

x(t)=37e2t+47e5t,y(t)=97e2t+27e5t.

Remark

An advantage of the fundamental matrix approach is this: Once we know the fundamental matrix Φ(t) and the inverse matrix Φ(0)1, we can calculate rapidly by matrix multiplication the solutions corresponding to different initial conditions. For example, suppose that we seek the solution of the system in (13) satisfying the new initial conditions x(0)=77, y(0)=49. Then substitution of (14) and (15) in (8) gives the new particular solution

x(t)=17[e2t2e5t3e2te5t][1231][7749]=17[e2t2e5t3e2te5t][21280]=[3e2t+80e5t9e2t+40e5t].

Exponential Matrices

We now discuss the possibility of constructing a fundamental matrix for the constant-coefficient linear system x=Ax directly from the coefficient matrix A-that is, without first applying the methods of earlier sections to find a linearly independent set of solution vectors.

We have seen that exponential functions play a central role in the solution of linear differential equations and systems, ranging from the scalar equation x=kx with solution x(t)=x0ekt to the vector solution x(t)=veλt of the linear system x=Ax whose coefficient matrix A has eigenvalue λ with associated eigenvector v. We now define exponentials of matrices in such a way that

X(t)=eAt

is a matrix solution of the matrix differential equation

X=AX

with n×n coefficient matrix A—in analogy with the fact that the ordinary exponential function x(t)=eat is a scalar solution of the first-order differential equation x=ax.

The exponential ez of the complex number z may be defined (as in Section 5.3) by means of the exponential series

(16)ez=1+z+z22!+z33!++znn!+.

Similarly, if A is an n×n matrix, then the exponential matrix eA is the n×n matrix defined by the series

(17)eA=I+A+A22!++Ann!+,

where I is the identity matrix. The meaning of the infinite series on the right in (17) is given by

(18)n=0Ann!=limk(n=0kAnn!),

where A0=I, A2=AA, A3=AA2, and so on; inductively, An+1=AAn if n0. It can be shown that the limit in (18) exists for every n×n square matrix A. That is, the exponential matrix eA is defined (by Eq. (17)) for every square matrix A.

Example 2

Consider the 2×2 diagonal matrix

A=[a00b].

Then it is apparent that

An=[an00bn]

for each integer n1. It therefore follows that

eA=I+A+A22!+=[1001]+[a00b]+[a2/2!00b2/2!]+=[1+a+a2/2!+001+b+b2/2!+].

Thus

eA=[ea00eb],

so the exponential of the diagonal 2×2 matrix A is obtained simply by exponentiating each diagonal element of A.

The n×n analog of the 2×2 result in Example 2 is established in the same way. The exponential of the n×n diagonal matrix

(19)D=[a1000a2000an]

is the n×n diagonal matrix

(20)eD=[ea1000ea2000ean],

obtained by exponentiating each diagonal element of D.

The exponential matrix eA satisfies most of the exponential relations that are familiar in the case of scalar exponents. For instance, if 0 is the n×n zero matrix, then Eq. (17) yields

(21)e0=I,

the n×n identity matrix. In Problem 31 we ask you to show that a useful law of exponents holds for n×n matrices that commute:

(22)IfAB=BA,theneA+B=eAeB.

In Problem 32 we ask you to conclude that

(23)(eA)1=eA.

In particular, the matrix eA is nonsingular for every n×n matrix A (reminiscent of the fact that ez0 for all z). It follows from elementary linear algebra that the column vectors of eA are always linearly independent.

If t is a scalar variable, then substitution of At for A in Eq. (17) gives

(24)eAt=I+At+A2t22!++Antnn!+.

(Of course, At is obtained simply by multiplying each element of A by t.)

Example 3

If,

A=[034006000],

then

A2=[0018000000]andA3=[000000000],

so An=0 for n3. It therefore follows from Eq. (24) that

eAt=I+At+12A2t2=[100010001]+[034006000]t+12[0018000000]t2;

that is,

eAt=[13t4t+9t2016t001].

Remark

If An=0 for some positive integer n, then the exponential series in (24) terminates after a finite number of terms, so the exponential matrix eA (or eAt) is readily calculated as in Example 3. Such a matrix—with a vanishing power—is said to be nilpotent.

Example 4

If

A=[234026002],

then

A=[200020002]+[034006000]=D+B,

where D=2I is a diagonal matrix and B is the nilpotent matrix of Example 3. Therefore, (20) and (22) give

eAt=e(D+B)t=eDteBt=[e2t000e2t000e2t][13t4t+9t2016t001];

thus

eAt=[e2t3te2t(4t+9t2)e2t0e2t6te2t00e2t].

Matrix Exponential Solutions

It happens that term-by-term differentiation of the series in (24) is valid, with the result

ddt(eAt)=A+A2t+A3t22!+=A(I+At+A2t22!+);

that is,

(25)ddt(eAt)=AeAt,

in analogy to the formula Dt(ekt)=kekt from elementary calculus. Thus the matrix-valued function

X(t)=eAt

satisfies the matrix differential equation

X=AX.

Because the matrix eAt is nonsingular, it follows that the matrix exponential eAt is a fundamental matrix for the linear system x=Ax. In particular, it is the fundamental matrix X(t) such that X(0)=I. Therefore, Theorem 1 implies the following result.

Thus the solution of homogeneous linear systems reduces to the task of computing exponential matrices. Conversely, if we already know a fundamental matrix Φ(t) for the linear system x=Ax, then the facts that eAt=Φ(t)C (by Eq. (4′)) and eA·0=e0=I (the identity matrix) yield

(28)eAt=Φ(t)Φ(0)1.

So we can find the matrix exponential eAt by solving the linear system x=Ax.

Example 5

In Example 1 we found that the system x=Ax with

A=[4231]

has fundamental matrix

Φ(t)=[e2t2e5t3e2te5t]withΦ(0)1=17[1231].

Hence Eq. (28) gives

eAt=17[e2t2e5t3e2te5t][1231]=17[e2t+6e5t2e2t+2e5t3e2t+3e5t6e2t+e5t].

Example 6

Use an exponential matrix to solve the initial value problem

(29)x=[234026002]x,x(0)=[192939].

Solution

The coefficient matrix A in (29) evidently has characteristic equation (2λ)3=0 and thus the triple eigenvalue λ=2, 2, 2. It is easy to see that the eigenvector equation

(A2I)v=[034006000][abc]=[000]

has (to within a constant multiple) the single solution v=[100]T. Thus there is only a single eigenvector associated with the eigenvalue λ=2, and so we do not yet have the three linearly independent solutions needed for a fundamental matrix. But we note that A is the same matrix whose matrix exponential

eAt=[e2t3te2t(4t+9t2)e2t0e2t6te2t00e2t]

was calculated in Example 4. Hence, using Theorem 2, the solution of the initial value problem in (29) is given by

x(t)=eAtx(0)=[e2t3te2t(4t+9t2)e2t0e2t6te2t00e2t][192939]=[(19+243t+351t2)e2t(29+234t)e2t39e2t].

Remark

The same particular solution x(t) as in Example 6 could be found using the generalized eigenvector method of Section 7.6. One would start by finding the chain of generalized eigenvectors

v1=[1800],v2=[460],v3=[001]

corresponding to the triple eigenvalue λ=2 of the matrix A. Then one would—using Eqs. (27) in Section 7.6—assemble the linearly independent solutions

x1(t)=v1e2t,x2(t)=(v1t+v2)e2t,x3(t)=(12v1t2+v2t+v3)e2t

of the differential equation x=Ax in (29). The final step would be to determine values of the coefficients c1, c2, c3 so that the particular solution x(t)=c1x1(t)+c2x2(t)+c3x3(t) satisfies the initial condition in (29). At this point it should be apparent that—especially if the matrix exponential eAt is readily available (for instance, from a computer algebra system)—the method illustrated in Example 6 can well be more “computationally routine” than the generalized eigenvector method.

General Matrix Exponentials

The relatively simple calculation of eAt carried out in Example 4 (and used in Example 6) was based on the observation that if

A=[234026002],

then A2I is nilpotent:

(30)(A2I)3=[034006000]3=[000000000]=0.

A similar result holds for any 3×3 matrix A having a triple eigenvalue r, in which case its characteristic equation reduces to (λr)3=0. For such a matrix, an explicit computation similar to that in Eq. (30) will show that

(31)(ArI)3=0.

(This particular result is a special case of the Cayley-Hamilton theorem of Section 6.3, according to which every matrix satisfies its own characteristic equation.) Thus the matrix ArI is nilpotent, and it follows that

(32)eAt=e(rI+ArI)t=erIte(ArI)t=ertI[I+(ArI)t+12(ArI)2t2],

the exponential series here terminating because of Eq. (31). In this way, we can rather easily calculate the matrix exponential eAt for any square matrix having only a single eigenvalue.

The calculation in Eq. (32) motivates a method of calculating eAt for any n×n matrix A whatsoever. As we saw in Section 7.6, A has n linearly independent generalized eigenvectors u1, u2, , un. Each generalized eigenvector u is associated with an eigenvalue λ of A and has a rank r1 such that

(33)(AλI)ru=0but(AλI)r1u0.

(If r=1, then u is an ordinary eigenvector such that Au=λu.)

Even if we do not yet know eAt explicitly, we can consider the function x(t)=eAtu, which is a linear combination of the column vectors of eAt and is therefore a solution of the linear system x=Ax with x(0)=u. Indeed, we can calculate x explicitly in terms of A, u, λ, and r:

x(t)=eAtu=e(λI+AλI)tu=eλIte(AλI)tu=eλtI[I+(AλI)t++(AλI)r1tr1(r1)!+]u,

so

(34)x(t)=eλt[u+(AλI)ut+(AλI)2ut22!++(AλI)r1utr1(r1)!],

using (33) and the fact that eλIt=eλtI.

If the linearly independent solutions x1(t), x2(t), , xn(t) of x=Ax are calculated using (34) with the linearly independent generalized eigenvectors u1, u2, , un, then the n×n matrix

(35)Φ(t)=[x1(t)x2(t)xn(t)]

is a fundamental matrix for the system x=Ax. Finally, the specific fundamental matrix X(t)=Φ(t)Φ(0)1 satisfies the initial condition X(0)=I, and thus is the desired matrix exponential eAt. We have therefore outlined a proof of the following theorem.

Example 7

Find eAt if

(37)A=[345054003].

Solution

Theorem 3 would apply even if the matrix A were not upper triangular. But because A is upper triangular, this fact enables us to see quickly that its characteristic equation is

(5λ)(3λ)2=0.

Thus A has the distinct eigenvalue λ1=5 and the repeated eigenvalue λ2=3.

Case 1: λ1=5. The eigenvector equation (AλI)u=0 for u=[abc]T is

(A5I)u=[245004002][abc]=[000].

The last two scalar equations 4c=0 and 2c=0 give c=0. Then the first equation 2a+4b=0 is satisfied by a=2 and b=1. Thus the eigenvalue λ1=5 has the (ordinary) eigenvector u1=[210]T. The corresponding solution of the system x=Ax is

(38)x1(t)=e5tu1=e5t[210]T.

Case 2: λ2=3. The eigenvector equation (AλI)u=0 for u=[abc]T is

(A3I)u=[045024000][abc]=[000].

The first two equations 4b+5c=0 and 2b+4c=0 imply that b=c=0, but leave a arbitrary. Thus the eigenvalue λ2=3 has the single (ordinary) eigenvector u2=[100]T. The corresponding solution of the system x=Ax is

(39)x2(t)=e3tu2=e3t[100]T.

To look for a generalized eigenvector of rank r=2 in Eq. (33), we consider the equation

(A3I)2u=[0816048000][abc]=[000].

The first two equations 8b+16c=0 and 4b+8c=0 are satisfied by b=2 and c=1, but leave a arbitrary. With a=0 we get the generalized eigenvector u3=[021]T of rank r=2 associated with the eigenvalue λ=3. Because (A3I)2u=0, Eq. (34) yields the third solution

(40)x3(t)=e3t [u3+(A3I)u3t]=e3t([021]+[045024000] [021] t)=e3t[3t21].

With the solutions listed in Eqs. (39) and (40), the fundamental matrix

Φ(t)=[x1(t)x2(t)x3(t)]

defined by Eq. (35) is

Φ(t)=[2e5te3t3te3te5t02e3t00e3t]withΦ(0)1=[012124001].

Hence Theorem 3 finally yields

eAt=Φ(t)Φ(0)1=[2e3te3t3te3te5t02e3t00e3t] [012124001]=[e3t2e5t2e3t4e5t(4+3t)e3t0e5t2e5t2e3t00e3t].

Remark

As in Example 7, Theorem 3 suffices for the computation of eAt provided that a basis consisting of generalized eigenvectors of A can be found. Alternatively, a computer algebra system can be used as indicated in the project material for this section.

8.1 Problems

Find a fundamental matrix of each of the systems in Problems 1 through 8, then apply Eq. (8) to find a solution satisfying the given initial conditions.

  1. x=[2112]x,x(0)=[32]

  2. x=[2142]x,x(0)=[21]

  3. x=[2542]x,x(0)=[01]

  4. x=[3111]x,x(0)=[10]

  5. x=[3293]x,x(0)=[11]

  6. x=[7543]x,x(0)=[20]

  7. x=[506212424]x,x(0)=[210]

  8. x=[322542553]x,x(0)=[101]

Compute the matrix exponential eAt for each system x=Ax given in Problems 9 through 20.

  1. x1=5x14x2, x2=2x1x2

  2. x1=6x16x2, x2=4x14x2

  3. x1=5x13x2, x2=2x1

  4. x1=5x14x2, x2=3x12x2

  5. x1=9x18x2, x2=6x15x2

  6. x1=10x16x2, x2=12x17x2

  7. x1=6x110x2, x2=2x13x2

  8. x1=11x115x2, x2=6x18x2

  9. x1=3x1+x2, x2=x1+3x2

  10. x1=4x1+2x2, x2=2x1+4x2

  11. x1=9x1+2x2, x2=2x1+6x2

  12. x1=13x1+4x2, x2=4x1+7x2

In Problems 21 through 24, show that the matrix A is nilpotent and then use this fact to find (as in Example 3) the matrix exponential eAt.

  1. A=[1111]

  2. A=[6496]

  3. A=[111111000]

  4. A=[303507303]

Each coefficient matrix A in Problems 25 through 30 is the sum of a nilpotent matrix and a multiple of the identity matrix. Use this fact (as in Example 6) to solve the given initial value problem.

  1. x=[2502] x,x(0)=[47]

  2. x=[70117] x,x(0)=[510]

  3. x=[123012001] x,x(0)=[456]

  4. x=[500105020305] x,x(0)=[405060]

  5. x=[1234016300120001] x,x(0)=[1111]

  6. x=[30006300963012963] x,x(0)=[1111]

  7. Suppose that the n×n matrices A and B commute; that is, that AB=BA. Prove that eA+B=eAeB. (Suggestion: Group the terms in the product of the two series on the right-hand side to obtain the series on the left.)

  8. Deduce from the result of Problem 31 that, for every square matrix A, the matrix eA is nonsingular with (eA)1=eA.

  9. Suppose that

    A=[0110].

    Show that A2n=I and that A2n+1=A if n is a positive integer. Conclude that

    eAt=I cosh t+A sinh t,

    and apply this fact to find a general solution of x=Ax. Verify that it is equivalent to the general solution found by the eigenvalue method.

  10. Suppose that

    A=[0220].

    Show that eAt=Icos 2t+12Asin 2t. Apply this fact to find a general solution of x=Ax, and verify that it is equivalent to the solution found by the eigenvalue method.

Apply Theorem 3 to calculate the matrix exponential eAt for each of the matrices in Problems 35 through 40.

  1. A=[3403]

  2. A=[123014001]

  3. A=[234013001]

  4. A=[5203001020005]

  5. A=[1333013300230002]

  6. A=[2444024400240003]

8.1 Application Automated Matrix Exponential Solutions

If A is an n×n matrix, then a computer algebra system can be used first to calculate the fundamental matrix eAt for the system

(1)X=Ax,

then to calculate the matrix product x(t)=eAtx0 to obtain a solution satisfying the initial condition x(0)=x0. For instance, suppose that we want to solve the initial value problem

x1=13x1+4x2,x2=4x1+7x2;x1(0)=11,x2(0)=23.

After the matrices

(2)A=[13447], x0=[1123]

have been entered, either the Maple command


with(linalg): exponential(A∗t)

the Mathematica command


MatrixExp[A t]

or the Matlab command


syms t, expm(A∗t)

yields the matrix exponential

expAt=15[e5t+4e15t2e5t+2e15t2e5t+2e15t4e5t+e15t].

Then either the Maple product multiply(expAt,x0), the Mathematica product expAt.x0, or the Matlab product expAt∗x0 gives the solution vector

x=[7e5t+18e15t14e5t+9e15t].

Obviously this, finally, is the way to do it!

Matrix exponentials also allow for convenient interactive exploration of the system (1). For example, a version of the Mathematica commands


A = {{13, 4}, {4, 7}};
field = VectorPlot[A.{x, y}, {x, −25, 25},
 {y, −25, 25}];
Manipulate[
 curves = ParametricPlot[
 MatrixExp[A t, #]&/@pt, {t, −1, 1},
 PlotRange −> 25];
 Show[curves, field],
 {{pt, {{11, 23}, {20, −10}, {−20, −10},
 {−20, 10}}}, Locator}]

was used to generate Fig. 8.1.1, which shows four solution curves of the system (1) with the matrix A chosen as in Eq. (2). The initial conditions for each solution curve-one of which initially passes through the point (11, 23) of our initial value problem, while the other three pass through the points (20,10), (20,10), and (20, 10)—are specified by a “locator point” which can be freely dragged to any desired position in the phase plane, with the corresponding solution curve being instantly redrawn.

FIGURE 8.1.1.

Interactive display of the linear system (1). As the “locator points” are dragged to different positions, the solution curves are immediately redrawn, illustrating the behavior of the system.

Experimenting with such interactive displays can shed considerable light on the behavior of linear systems. For example, notice the straight line solution in Fig. 8.1.1; if you could drag the corresponding locator point around the phase plane, what other straight line solution could you find?How could you have predicted this by examining the matrix A?

For a three-dimensional example, solve the initial value problem

x1=149x150x2154x3,x2=537x1+180x2+546x3,x3=27x19x225x3;x1(0)=17,x2(0)=43,x3(0)=79.

And here’s a four-dimensional problem:

x1=4x1+x2+x3+7x4,x2=x1+4x2+10x3+x4,x3=x1+10x2+4x3+x4,x4=7x1+x2+x3+4x4;x1(0)=15,x2(0)=35,x3(0)=55,x4(0)=75.

If at this point you’re having too much fun with matrix exponentials to stop, make up some problems of your own. For instance, choose any homogeneous linear system appearing in this chapter and experiment with different initial conditions. The exotic 5×5 matrix A of the Section 7.6 application may suggest some interesting possibilities.

8.2 Nonhomogeneous Linear Systems

In Section 5.5 we exhibited two techniques for finding a single particular solution of a single nonhomogeneous nth-order linear differential equation—the method of undetermined coefficients and the method of variation of parameters. Each of these may be generalized to nonhomogeneous linear systems. In a linear system modeling a physical situation, nonhomogeneous terms typically correspond to external influences, such as the inflow of liquid to a cascade of brine tanks or an external force acting on a mass-and-spring system.

Given the nonhomogeneous first-order linear system

(1)x=Ax+f(t)

where A is an n×n constant matrix and the “nonhomogeneous term” f(t) is a given continuous vector-valued function, we know from Theorem 4 of Section 7.2 that a general solution of Eq. (1) has the form

(2)x(t)=xc(t)+xp(t),

where

Preceding sections have dealt with xc(t), so our task now is to find xp(t).

Undetermined Coefficients

First we suppose that the nonhomogeneous term f(t) in (1) is a linear combination (with constant vector coefficients) of products of polynomials, exponential functions, and sines and cosines. Then the method of undetermined coefficients for systems is essentially the same as for a single linear differential equation. We make an intelligent guess as to the general form of a particular solution xp, then attempt to determine the coefficients in xp by substitution in Eq. (1). Moreover, the choice of this general form is essentially the same as in the case of a single equation (discussed in Section 5.5); we modify it only by using undetermined vector coefficients rather than undetermined scalars. We will therefore confine the present discussion to illustrative examples.

Example 1

Find a particular solution of the nonhomogeneous system

(3)x=[3275] x+[32t].

Solution

The nonhomogeneous term f=[32t]T is linear, so it is reasonable to select a linear trial particular solution of the form

(4)xp(t)=at+b=[a1a2] t+[b1b2].

Upon substitution of x=xp in Eq. (3), we get

[a1a2]=[3275] [a1t+b2a2t+b2]+[32t]=[3a1+2a27a1+5a2+2] t+[3b1+2b2+37b1+5b2].

We equate the coefficients of t and the constant terms (in both x1- and x2-components) and thereby obtain the equations

(5)3a1+2a2=0,7a1+5a2+2=0,3b1+2b2+3=a1,7b1+5b2=a2,

We solve the first two equations in (5) for a1=4 and a2=6. With these values we can then solve the last two equations in (5) for b1=17 and b2=25. Substitution of these coefficients in Eq. (4) gives the particular solution x=[x1x2]T of (3) described in scalar form by

x1(t)=4t+17,x2(t)=6t25.

Example 2

Cascading brine tanks Figure 8.2.1 shows the system of three brine tanks investigated in Example 2 of Section 7.3. The volumes of the three tanks are V1=20,=V2=40, and V3=50(gal), and the common flow rate is r=10 (gal/min). Suppose that all three tanks contain fresh water initially, but that the inflow to tank 1 is brine containing 2 pounds of salt per gallon, so that 20 pounds of salt flow into tank 1 per minute. Referring to Eq. (18) in Section 7.3, we see that the vector x(t)=[x1(t)x2(t)x3(t)]T of amounts of salt (in pounds) in the three tanks at time t satisfies the nonhomogeneous initial value problem

(6)dxdt=[0.5000.50.25000.250.2] x+[2000]x(0)=[000].

FIGURE 8.2.1.

The three brine tanks of Example 2.

The nonhomogeneous term f=[2000]T here corresponds to the 20 lb/min inflow of salt to tank 1, with no (external) inflow of salt into tanks 2 and 3.

Because the nonhomogeneous term is constant, we naturally select a constant trial function xp=[a1a2a3]T, for which xp0. Then substitution of x=xp in (6) yields the system

[000]=[0.5000.50.25000.250.2] [a1a2a3]+[2000]

that we readily solve for a1=40, a2=80, and a3=100 in turn. Thus our particular solution is xp(t)=[4080100]T.

In Example 2 of Section 7.3 we found the general solution

xc(t)=c1 [365] et/2+c2[015] et/4+c3[001] et/5

of the associated homogeneous system, so a general solution x=xc+xp of the nonhomogeneous system in (6) is given by

(7)x(t)=c1[365] et/2+c2 [015] et/4+c3 [001] et/5+[4080100].

When we apply the zero initial conditions in (6), we get the scalar equations

3c1+40=0,6c1+c2+80=0,5c15c2+c3+100=0

that are readily solved for c1=403, c2=160, and c3=25003. Substituting these coefficients in Eq. (7), we find that the amounts of salt in the three tanks at time t are given by

(8)x1(t)=4040et/2,x2(t)=80+80et/2160et/4x3(t)=100+1003(2et/2+24et/425et/5).

As illustrated in Fig. 8.2.2, we see the salt in each of the three tanks approaching, as t+, a uniform density of 2 lb/gal—the same as the salt density in the inflow to tank 1.

FIGURE 8.2.2.

Solution curves for the amount of salt defined in (8).

In the case of duplicate expressions in the complementary function and the nonhomogeneous terms, there is one difference between the method of undetermined coefficients for systems and for single equations (Rule 2 in Section 5.5). For a system, the usual first choice for a trial solution must be multiplied not only by the smallest integral power of t that will eliminate duplication, but also by all lower (nonnegative integral) powers of t as well, and all the resulting terms must be included in the trial solution.

Example 3

Consider the nonhomogeneous system

(9)x=[4231] x[154] te2t.

In Example 1 of Section 7.3 we found the solution

(10)xc(t)=c1 [13] e2t+c2 [21] e5t

of the associated homogeneous system. A preliminary trial solution xp(t)=ate2t+be2t exhibits duplication with the complementary function in (10). We would therefore select

xp(t)=at2e2t+bte2t+ce2t

as our trial solution, and we would then have six scalar coefficients to determine. It is simpler to use the method of variation of parameters, our next topic.

Variation of Parameters

Recall from Section 5.5 that the method of variation of parameters may be applied to a linear differential equation with variable coefficients and is not restricted to nonhomogeneous terms involving only polynomials, exponentials, and sinusoidal functions. The method of variation of parameters for systems enjoys the same flexibility and has a concise matrix formulation that is convenient for both practical and theoretical purposes.

We want to find a particular solution xp of the nonhomogeneous linear system

(11)x=P(t)x+f(t),

given that we have already found a general solution

(12)xc(t)=c1x1(t)+c2x2(t)++cnxn(t)

of the associated homogeneous system

(13)x=P(t)x.

We first use the fundamental matrix Φ(t) with column vectors x1, x2, , xn to rewrite the complementary function in (12) as

(14)xc(t)=Φ(t)c,

where c denotes the column vector whose entries are the coefficients c1,c2,,cn. Our idea is to replace the vector “parameter” c with a variable vector u(t). Thus we seek a particular solution of the form

(15)xp(t)=Φ(t)u(t).

We must determine u(t) so that xp does, indeed, satisfy Eq. (11).

The derivative of xp(t) is (by the product rule)

(16)xp(t)=Φ(t)u(t)+Φ(t)u(t).

Hence substitution of Eqs. (15) and (16) in (11) yields

(17)Φ(t)u(t)+Φ(t)u(t)=P(t)Φ(t)u(t)+f(t).

But

(18)Φ(t)=P(t)Φ(t)

because each column vector of Φ(t) satisfies Eq. (13). Therefore, Eq. (17) reduces to

(19)Φ(t)u(t)=f(t).

Thus it suffices to choose u(t) so that

(20)u(t)=Φ(t)1f(t);

that is, so that

(21)u(t)=Φ(t)1f(t) dt.

Upon substitution of (21) in (15), we finally obtain the desired particular solution, as stated in the following theorem.

This is the variation of parameters formula for first-order linear systems. If we add this particular solution and the complementary function in (14), we get the general solution

(23)x(t)=Φ(t)c+Φ(t) Φ(t)1f(t) dt

of the nonhomogeneous system in (11).

The choice of the constant of integration in Eq. (22) is immaterial, for we need only a single particular solution. In solving initial value problems it often is convenient to choose the constant of integration so that xp(a)=0, and thus integrate from a to t:

(24)xp(t)=Φ(t) atΦ(s)1f(s) ds.

If we add the particular solution of the nonhomogeneous problem

x=P(t)x+f(t),x(a)=0

in (24) to the solution xc(t)=Φ(t)Φ(a)1xa of the associated homogeneous problem x=P(t)x, x(a)=xa, we get the solution

(25)x(t)=Φ(t)Φ(a)1xa+Φ(t) atΦ(s)1f(s) ds

of the nonhomogeneous initial value problem

(26)x=P(t)x+f(t),x(a)=xa.

Equations (22) and (25) hold for any fundamental matrix Φ(t) of the homogeneous system x=P(t)x. In the constant-coefficient case P(t)A we can use for Φ(t) the exponential matrix eAt—that is, the particular fundamental matrix such that Φ(0)=I. Then, because (eAt)1=eAt, substitution of Φ(t)=eAt in (22) yields the particular solution

(27)xp(t)=eAt eAtf(t) dt

of the nonhomogeneous system x=P(t)x+f(t). Similarly, substitution of Φ(t)=eAt in Eq. (25) with a=0 yields the solution

(28)x(t)=eAtx0+eAt oteAtf(t) dt

of the initial value problem

(29)x=P(t)x+f(t),x(0)=x0.

Remark

If we retain t as the independent variable but use s for the variable of integration, then the solutions in (27) and (28) can be rewritten in the forms

xp(t)=eA(st)f(s) dsandx(t)=eAtx0+0teA(st)f(s) ds.

Example 4

Solve the initial value problem

(30)x=[4231] x[154] te2t,x(0)=[73].

Solution

The solution of the associated homogeneous system is displayed in Eq. (10). It gives the fundamental matrix

Φ(t)=[e2t2e5t3e2te5t]withΦ(0)1=17[1231].

It follows by Eq. (28) in Section 8.1 that the matrix exponential for the coefficient matrix A in (30) is

eAt=Φ(t)Φ(0)1=[e2t2e5t3e2te5t]17[1231]=17[e2t+6e5t2e2t+2e5t3e2t+3e5t6e2t+e5t].

Then the variation of parameters formula in Eq. (28) gives

eAtx(t)=x0+0teAsf(s) ds=[73]+0t17[e2s+6e5s2e2s+2e5s32s+3e5s6e2s+e5s][15se2s4se2s] ds=[73]+0t[s14se7s3s7se7s] ds=[73]+114[47t2+4e7t+28te7t2+21t2+2e7t+14te7t].

Therefore,

eAtx(t)=114[947t2+4e7t+28te7t40+21t2+2e7t+14te7t].

Upon multiplication of the right-hand side here by eAt, we find that the solution of the initial value problem in (30) is given by

x(t)=17[e2t+6e5t2e2t+2e5t3e2t+3e5t6e2t+e5t]114[947t2+4e7t+28te7t40+21t2+2e7t+14te7t]=114[(6+28t7t2)e2t+92e5t(4+14t+21t2)e2t+46e5t].

In conclusion, let us investigate how the variation of parameters formula in (22) “reconciles” with the variation of parameters formula in Theorem 1 of Section 5.5 for the second-order linear differential equation

(31)y+Py+Qy=f(t).

If we write y=x1, y=x1=x2, y=x1=x2, then the single equation in (31) is equivalent to the linear system x1=x2, x2=Qx1Px2+f(t), that is,

(32)x=P(t)x+f(t),

where

x=[x1x2]=[yy],P(t)=[01QP],andf(t)=[0f(t)].

Now two linearly independent solutions y1 and y2 of the homogeneous system y+Py+Qy=0 associated with (31) provide two linearly independent solutions

x1=[y1y1]andx2=[y2y2]

of the homogeneous system x=P(t)x associated with (32). Observe that the determinant of the fundamental matrix Φ=[x1x2] is simply the Wronskian

W=|y1y2y1y2|

of the solutions y1 and y2, so the inverse fundamental matrix is

Φ1=1W|y2y2y1y1|.

Therefore the variation of parameters formula xp=ΦΦ1f dt in (22) yields

[ypyp]=[y1y2y1y2]1W[y2y2y1y1][0f]dt=[y1y2y1y2]1W[y2fy1f]dt.

The first component of this column vector is

yp=[y1y2]1W[y2fy1f]dt=y1y2fWdt+y2y1fW dt.

If, finally, we supply the independent variable t throughout, the final result on the right-hand side here is simply the variation of parameters formula in Eq. (33) of Section 5.5 (where, however, the independent variable is denoted by x).

8.2 Problems

Apply the method of undetermined coefficients to find a particular solution of each of the systems in Problems 1 through 14. If initial conditions are given, find the particular solution that satisfies these conditions. Primes denote derivatives with respect to t.

  1. x=x+2y+3, y=2x+y2

  2. x=2x+3y+5, y=2x+y2t

  3. x=3x+4y, y=3x+2y+t2;x(0)=y(0)=0

  4. x=4x+y+et, y=6xyet;x(0)=y(0)=1

  5. x=6x7y+10, y=x2y2et

  6. x=9x+y+2et, y=8x2y+tet

  7. x=3x+4y+sin t, y=6x5y;x(0)=1, y(0)=0

  8. x=x5y+2sin t, y=xy3cos t

  9. x=x5y+cos 2t, y=xy

  10. x=x2y, y=2xy+et sin t

  11. x=2x+4y+2, y=x+2y+3;x(0)=1, y(0)=1

  12. x=x+y+2t, y=x+y2t

  13. x=2x+y+2et, y=x+2y3et

  14. x=2x+y+1, y=4x+2y+e4t

Two Brine Tanks

Problems 15 and 16 are similar to Example 2, but with two brine tanks (having volumes V1 and V2 gallons as in Fig. 8.2.1) instead of three tanks. Each tank initially contains fresh water, and the inflow to tank 1 at the rate of r gallons per minute has a salt concentration of c0 pounds per gallon. (a) Find the amounts x1(t) and x2(t) of salt in the two tanks after t minutes. (b) Find the limiting (long-term) amount of salt in each tank. (c) Find how long it takes for each tank to reach a salt concentration of 1 lb/gal.

  1. V1=100, V2=200, r=10, c0=2

  2. V1=200, V2=100, r=10, c0=3

In Problems 17 through 34, use the method of variation of parameters (and perhaps a computer algebra system) to solve the initial value problem

x=Ax+f(t),x(a)=xa.

In each problem we provide the matrix exponential eAt as provided by a computer algebra system.

  1. A=[6712], f(t)=[6090], x(0)=[00],eAt=16[et+7e5t7et7e5tet+e5t7ete5t]

  2. Repeat Problem 17, but with f(t) replaced with [100t50t].

  3. A=[1222], f(t)=[180t90], x(0)=[00],eAt=15[e3t+4e2t2e3t+2e2t2e3t+2e2t4e3t+e2t] 

  4. Repeat Problem 19, but with f(t) replaced with [75e2t0].

  5. A=[4152], f(t)=[18e2t30e2t], x(0)=[00],eAt=14[et+5e3tete3t5et+5e5t5ete3t]

  6. Repeat Problem 21, but with f(t) replaced with [28et20e3t].

  7. A=[3193],f(t)=[75], x(0)=[35],eAt=[1+3tt9t13t]

  8. Repeat Problem 23, but with f(t)=[0t2] and x(1)=[37].

  9. A=[2512],f(t)=[4t1], x(0)=[00],eAt=[cos t+2 sin t5 sin tsin tcos2 sin t]

  10. Repeat Problem 25, but with f(t)=[4 cos t6 sin t] and x(0)=[35].

  11. A=[2412], f(t)=[36t26t], x(0)=[00],eAt=[1+2t4tt12t]

  12. Repeat Problem 27, but with f(t)=[4ln tt1] and x(1)=[11].

  13. A=[0110], f(t)=[sec t0], x(0)=[00],eAt=[cos tsin tsin tcos t]

  14. A=[0220], f(t)=[t cos 2tt sin 2t], x(0)=[00],eAt=[cos 2tsin 2tsin 2tcos 2t]

  15. A=[123012001], f(t)=[006et], x(0)=[000],eAt=[et2tet(3t+2t2)et0et2tet00et]

  16. A=[134013002], f(t)=[002e2t], x(0)=[000],eAt=[et3tet(139t)et+13e2t0et3et+3e2t00e2t]

  17. A=[0430003800040000], f(t)=30 [tttt], x(0)=[0000],eAt=[14t8t+6t232t2+8t2013t8t+6t20014t0001]

  18. A=[0480000800240002], f(t)=[06t0e2t], x(0)=[4221],eAt=[14t4(1+e2t)16t(1+e2t)0104(1+e2t)00e2t4te2t000e2t]

8.2 Application Automated Variation of Parameters

The application of the variation of parameters formula in Eq. (28) encourages so mechanical an approach as to encourage especially the use of a computer algebra system. The following Mathematica commands were used to check the results in Example 4 of this section.


A = {{4,2}, {3,−1}};
x0 = {{7}, {3}};
f[t_] := {{−15 t Exp[−2t]},{−4 t Exp[−2t]}};
exp[A_] := MatrixExp[A]
x = exp[A∗t].(x0 + Integrate[exp[−A∗s].f[s], {s,0,t}])

The matrix exponential commands illustrated in the Section 5.6 application provide the basis for analogous Maple and Matlab computations. You can then check routinely the answers for Problems 17 through 34 of this section.

8.3 Spectral Decomposition Methods

Here, we present an alternative approach to the computation of the matrix exponential eAt, one that does not require that eigenvectors (including generalized ones) of the n×n matrix A be found first. Assume that the characteristic polynomial of A is written in the form

(1)p(λ)=(1)n|AλI|,

with leading term +λn. [Compare Eqs. (4) and (5) in Section 6.1.] If the (not necessarily distinct) eigenvalues of A are λ1,λ2,,λn, then

(2)p(λ)=(λλ1)(λλ2)(λλn).

The Cayley-Hamilton theorem (Section 6.3) says that any matrix A satisfies its own characteristic equation; that is,

(3)p(A)=i=1n(AλiI)=0

(where I denotes the n×n identity matrix). This crucial fact is the key to our method in this section.

The way we proceed to calculate the matrix exponential eAt depends on whether or not the eigenvalues of A are distinct.

The Case of Distinct Eigenvalues

If the eigenvalues of A are distinct (whether real or complex), the reciprocal 1/p(λ) has a partial fraction decomposition of the form

(4)1p(λ)=a1λλ1+a2λλ2++anλλn,

where the numerators a1,a2,,an are constants. Indeed, these constants can be found by multiplying Eq. (4) by p(λ) to get

(5)1=i=1naibi(λ)=a1b1(λ)+a2b2(λ)++anbn(λ),

where the polynomial

(6)bi(λ)=p(λ)λλi=ji(λλj)

is obtained from the characteristic polynomial p(λ) upon deletion of the factor (λλi) corresponding to the ith eigenvalue.

It follows immediately from (6) that bi(λi)0, whereas bj(λi)=0 if ij [because in this case bj(λi) includes the factor (λiλi)]. Consequently, substitution of λ=λi into Eq. (5) gives the value

(7)ai=1bi(λi)=1ji(λiλj)

for i=1,2,,n. Given these coefficient values, we define the projection matrices P1,P2,,Pn of the fixed n×n matrix A by writing

(8)Pi=aibi(A)=aiji(AλjI)=ji(AλjI)ji(λiλj).

These matrices have very special properties that are summarized by the following proposition.

Remark

In terms of the Kronecker delta defined by

δij={1if i=j0if ij,

the conditions in (10) and (11) say that

(13)PiPj=δijPi.

Also, note that (12) says that Pi is effectively an “eigenmatrix” associated with the eigenvalue λi of the matrix A—that is, multiplication of Pi by the matrix A (on the right) gives a scalar multiple of Pi.

The following theorem expresses the matrix A in terms of its projection matrices.

Remark 1

If we square both sides in (14) and then use (13), we get

(λ1P1+λ2P2++λnPn)(λ1P1+λ2P2++λnPn)=i,j=1nλiλjPiPj,

or

A2=i=1nλi2Pi=λ12P2+λ22P2++λn2Pn.

Continuing in this fashion, we deduce by repeated multiplication that

(15)Ak=i=1nλikPi=λ1kP1+λ2kP2++λnkPn

for any positive integer k. We will see in the proof of Theorem 2 that this single fact is all we need to express the matrix exponential eAt simply and explicitly in terms of the eigenvalues and projection matrices of the matrix A.

Remark 2

The relation in (15) holds also if k=12. For instance, if n=2 and B=λ1P1+λ2P2, then the basic properties of projection matrices (Proposition 1) and (14) yield

B2=(λ1P1+λ2P2)(λ1P1+λ2P2)=λ1P12+λ1λ2P1P2+λ2λ1P2P1+λ2P22=λ1P1+λ2P2=A.

It follows that if (for each i) λi denotes either square root of the eigenvalue λi of A, then

(16)A=λ1P1+λ2P2

is a square root of the matrix A.

Example 1

If A is a 2×2 matrix with distinct eigenvalues λ1 and λ2, then Eqs. (7), (8), and (17) give first

(18)a1=1λ1λ2anda2=1λ2λ1,

then

(19)P1=Aλ2Iλ1λ2andP2=Aλ1Iλ2λ1,

and finally

(20)eAt=eλ1tP1+eλ2tP2.

Thus the calculation of the matrix exponential eAt using the projection matrices P1 and P2 reduces to a routine matter of numerical substitution.

Example 2

The matrix

A=[10138]

has the characteristic polynomial

p(λ)=(10λ)(8λ)3=λ218λ+77=(λ7)(λ11),

so its eigenvalues are λ1=7 and λ2=11. Hence Eq. (19) gives

P1=14[101113811]=14[1133]

and

P2=14[1071387]=14[3131],

so Eq. (20) gives

eAt=14e7t[1133]+14e11t[3131]=14[e7t+3e11te7t+e11t3e7t+3e11t3e7t+e11t].

Example 3

The 3×3 matrix

A=[30046216155]

has the characteristic polynomial

p(λ)=|AλI|=(3λ)[(6λ)(5λ)+30]=(3λ)[λ2λ]p(λ)=λ(λ1)(λ3).

Hence the eigenvalues of A are λ1=0, λ2=1, and λ3=3. To find the matrix exponential

(21)eAt=eλ1tP1+eλ2tP2+eλ3tP3,

we need only calculate the three projections matrices P1, P2, and P3 defined by Eq. (8). But first we need the coefficients

a1=1(λ1λ2)(λ1λ3)=1(01)(03)=13,a2=1(λ2λ1)(λ2λ3)=1(10)(13)=12,anda3=1(λ3λ1)(λ3λ2)=1(30)(31)=16

that are given by Eq. (7). Then

P1=a1(Aλ2I)(Aλ3I)=13[20045216156][00043216158]=[00045212156],P2=a2(Aλ1I)(Aλ3I)=12[30046216155][00043216158]=[00046210155],

and

P3=a3(Aλ1I)(Aλ2I)=16[30046216155][20045216156]=[100000200].

Finally, (17) gives the desired matrix exponential

eAt=e0t[00045212156]+e1t[00046210155]+e3t[100000200]=[e3t0044et5+6et2+2et12+10et+2e3t1515et65et].

As an application, the solution of the initial value problem

x(t)=Ax,x(0)=x0=[3711]T

is (by Theorem 2 in Section 8.1) given by

x(t)=eAtx0=[e3t0044et5+6et2+2et12+10et+2e3t1515et65et][3711]=[3e3t45+52et135130et+6e3t].

Second-Order Linear Systems

Spectral decompositions of matrices also yield convenient solutions of second-order matrix linear systems.

Example 4

Mass-spring system Consider the mass-and-spring system shown in Fig. 8.3.1, with m1=2, m2=1, k1=100, and k2=50. As in Example 1 of Section 7.4, the position vector x(t)=[x1(t)x2(t)]T satisfies the second-order system x=Ax having coefficient matrix

A=[75255050].

FIGURE 8.3.1.

The mass-and-spring system of Example 4.

We find that A has characteristic polynomial p(λ)=λ2+125λ+2500=(λ+25)(λ+100) and hence has eigenvalues λ1=25 and λ2=100. By using (19), we calculate the projection matrices

P1=Aλ2Iλ1λ2=175[25255050]=13[1122]andP2=Aλ1Iλ2λ1=175[50255025]=13[2121]

of A. Starting with the spectral decomposition

A=λ1P1+λ2P2=5iP1+10iP2

of A, Theorem 2 yields

eAt=e5itP1+e10itP2andeAt=e5itP1+e10itP2.

Consequently, Eq. (22) in Theorem 3 gives the following general solution:

(23)x(t)=eAtc1+eAtc2=(e5itP1+e10itP2)c1+(e5itP1+e10itP2)c2=P1(c1e5it+c2e5it)+P2(c1e10it+c2e10it)x(t)=P1(a cos 5t+b sin 5t)+P2(a cos 10t+b sin 10t),

where a=c1+c2 and b=i(c1c2). Now

p1a=13[1122][a1a2]=13(a1+a2)[12]andP1b=13(b1+b2)[12],P2a=13[2121][a1a2]=13(2a1a2)[11]

and

P2b=13(2b1b2)[11].

Hence (23) finally takes the form

(24)x(t)=(acos5t+bsin5t)[12]+(ccos10t+dsin10t)[11].

This last equation expresses the motion x(t) of the mass-spring system in Fig. 8.3.1 as a linear combination of two natural modes of free oscillation. In the first mode, the two masses move with frequency ω1=5 in the same direction and with the amplitude of m2 being twice that of m1. In the second mode, the two masses move with frequency ω2=10 in opposite directions and with equal amplitudes of motion.

The General Case

Now we want to take into account the possibility of multiple eigenvalues. If the n×n matrix A has distinct eigenvalues λ1,λ2,,λq having multiplicities m1,m2,,mq (respectively), then the reciprocal 1/p(λ) has a partial-fraction decomposition of the form

(25)1p(λ)=a1(λ)(λλ1)m1+a2(λ)(λλ2)m2++aq(λ)(λλq)mq,

where (for each i=1,2,,q) the numerator ai(λ) is a polynomial in λ of degree at most mi1. Once these numerator polynomials have been found, we can define the projection matrices P1,P2,,Pq of A by the formula

(26)Pi=ai(A)bi(A)=ai(A)Πji(AλjI)mj,

analogous to Eq. (8), with the polynomial

(27)bi(λ)=p(λ)(λλi)mi=Πji(λλj)mj

being obtained from the characteristic polynomial p(λ) upon deletion of the factor (λλi)mi corresponding to the ith eigenvalue. It follows immediately from (27)that bi(λi)0, whereas bj(λi)=0 if ij[in which case bj(λi) includes the factor (λiλi)mi]. The following properties of the projection matrices P1,P2,,Pq are established in essentially the same way as in the proof of Proposition 1.

In the distinct-eigenvalue case we saw in part (iv) of Proposition 1 that Pi(AλiI)=0 for each i. This may no longer be so if mi>1, so we define

(31)Ni=Pi(AλiI)

for each i=1,2,,q. Part (ii) of the following proposition asserts that these matrices are nilpotent. Note that Ni=0 if mi=1 (so that λi is a nonrepeated eigenvalue).

The following theorem expresses the matrix A in terms of its projection and nilpotent matrices in the general case (allowing multiple eigenvalues).

As in the distinct-eigenvalue case of Theorem 2, the spectral decomposition of Theorem 4 can be used to calculate the matrix exponential eAt in the general case.

The complicated formula in (35) looks much simpler in low-dimensional special cases.

Example 5

If A is a 2×2 matrix with a single eigenvalue λ1 of multiplicity 2 and corresponding projection matrix P1, then (35) reduces to eAt=eλ1tP1[I+(Aλ1I)t]. In this case, p(λ)=(λλ1)2 so Eqs. (25)–(27) give a1(λ)=b1(λ)=1, and hence P1=I. It therefore follows that

eAt=eλ1t[I+(Aλ1I)t].

Example 6

If A is a 3×3 matrix with an eigenvalue λ1 of multiplicity 1, an eigenvalue λ2 of multiplicity 2, and corresponding projection matrices P1 and P2, then (35) reduces to

eAt=eλ1tP1+eλ2tP2[I+(Aλ2I)t].

Example 7

If A is a 5×5 matrix with an eigenvalue λ1 of multiplicity 2 and an eigenvalue λ2 of multiplicity 3, then (35) reduces to

eAt=eλ1tP1[I+(Aλ1I)t]+eλ2tP2[I+(Aλ2I)t+12(Aλ2I)2t2].

In this case, the principal labor may consist of finding the characteristic polynomial and its partial-fractions decomposition (25), so that Eqs. (26)(27) can be used to find the projection matrices P1 and P2.

Example 8

The matrix

(37)A=[421201223]

has the characteristic polynomial

p(λ)=|AλI|=λ37λ2+16λ12=(λ3)(λ2)2

and thus has eigenvalues λ1=3 of multiplicity 1 and λ2=2 of multiplicity 2. The partial-fractions decomposition

1(λ3)(λ2)2=1λ3+1λ(λ2)2

shows that a1(λ)=1 and a2=1λ. Now b1(λ)=(λ2)2 and b2(λ)=λ3, so (26)gives

P1=a1(A)b1(A)=(A2I)2=[221221221]2=[221221221]

and

P2=a2(A)b2(A)=(IA)(A3I)=[321211222][121231220]=[121231220].

The result of Example 6 therefore gives

eAt=eλ1tP1+eλ2tP2[I+(Aλ2I)t]=e3t[221221221]+e2t[121231220]([100010001]+[221221221]t)eAt=[2e3te2t2e3t+2e2te3te2t2e3t2e2t2e3t+3e2te3te2t2e3t2e2t2e3t+2e2te3t].

For instance, suppose we want to solve the initial value problem

(38)x(t)=Ax+f,x(0)=x0,

where A is the matrix in (37), x0=[293743]T, and f=[6000]T. In this case, the variation of parameters formula in Eq. (28) of Section 8.2 gives

eAtx(t)=x0+0teAsf ds=[293743]+0t[2e3se2s2e3s+2e2se3se2s2e3s2e2s2e3s+3e2se3se2s2e3s2e2s2e3s+2e2se3s][6000] ds=[293743]+0t[120e3s60e2s120e3s120e2s120e3s120e2s] dseAtx(t)=[293743]+[1040e3t+30e2t2040e3t+60e2t2040e3t+60e2t]=[3940e3t+30e2t1740e3t+60e2t2340e3t+60e2t].

Finally, multiplication by eAt yields the solution

x(t)=[2e3te2t2e3t+2e2te3te2t2e3t2e2t2e3t+3e2te3te2t2e3t2e2t2e3t+2e2te3t][3940e3t+30e2t1740e3t+60e2t2340e3t+60e2t]=[10+67e3t28e2t20+67e3t50e2t20+67e3t44e2t]

of the initial value problem in (38).

8.3 Problems

  1. 1–20. Use projection matrices to find a fundamental matrix solution x(t)=eAt of each of the linear systems x=Ax given in Problems 1 through 20 of Section 7.3 .

  2. 21–30. Use projection matrices to find a fundamental matrix solution of each of the linear systems given in Problems 1 through 10 of Section 7.6 .

  3. 31–40. Use projection matrices, as in Example 8 of this section, to find the matrix exponentials and particular solutions desired in Problems 21 through 30 (respectively) of Section 8.2 .

In each of Problems 41 through 46, use the spectral decomposition methods of this section to find a fundamental matrix solution x(t)=eAt for the linear system x=Ax given in the problem.

Use projection matrices as in Example 4 of this section to solve Problems 47 through 50. You can simplify matters by taking c1=[10]T and c2=[01]T to calculate a typical particular solution x(t)=eAtc1+eAtc2 that exhibits the fundamental frequencies and modes of oscillation.

9 Nonlinear Systems and Phenomena

9.1 Stability and the Phase Plane

A wide variety of natural phenomena are modeled by two-dimensional first-order systems of the form

(1)dxdt=F(x,y),dydt=G(x,y)

in which the independent variable t does not appear explicitly. We usually think of the dependent variables x and y as position variables in the xy-plane and of t as a time variable. We will see that the absence of t on the right-hand sides in (1) makes the system easier to analyze and its solutions easier to visualize. Using the terminology of Section 2.2, such a system of differential equations in which the derivative values are independent (or “autonomous”) of time t is often called an autonomous system.

We generally assume that the functions F and G are continuously differentiable in some region R of the xy-plane. Then according to the existence and uniqueness theorems of Appendix A, given t0 and any point (x0, y0) of R, there is a unique solution x=x(t), y=y(t) of (1) that is defined on some open interval (a, b) containing t0 and satisfies the initial conditions

(2)x(t0)=x0,y(t0)=y0.

The equations x=x(t), y=y(t) then describe a parametrized solution curve in the phase plane. Any such solution curve is called a trajectory of the system in (1), and precisely one trajectory passes through each point of the region R (Problem 29). A critical point of the system in (1) is a point (x,y) such that

(3)F(x,y)=G(x,y)=0.

If (x,y) is a critical point of the system, then the constant-valued functions

(4)x(t)x,y(t)=y

have derivatives x(t)0 and y(t)0, and therefore automatically satisfy the equations in (1). Such a constant-valued solution is called an equilibrium solution of the system. Note that the trajectory of the equilibrium solution in (4) consists of the single point (x,y).

In some practical situations these very simple solutions and trajectories are the ones of greatest interest. For example, suppose that the system x=F(x,y), y=G(x,y) models two populations x(t) and y(t) of animals that cohabit the same environment, and perhaps compete for the same food or prey on one another; x(t) might denote the number of rabbits and y(t) the number of squirrels present at time t. Then a critical point (x,y) of the system specifies a constant population x of rabbits and a constant population y of squirrels that can coexist with one another in the environment. If (x0,y0) is not a critical point of the system, then it is not possible for constant populations of x0 rabbits and y0 squirrels to coexist; one or both must change with time.

Example 1

Find the critical points of the system

(5)dxdt=14x2x2xy,dydt=16y2y2xy.

Solution

When we look at the equations

14x2x2xy=x(142xy)=0,16y2y2xy=y(162yx)=0

that a critical point (x, y) must satisfy, we see that either

(6a)x=0or142xy=0,

and either

(6b)y=0or162yx=0.

If x=0 and y0, then the second equation in (6b) gives y=8. If y=0 and x0, then the second equation in (6a) gives x=7. If both x and y are nonzero, then we solve the simultaneous equations

2x+y=14,x+2y=16

for x=4, y=6. Thus the system in (5) has the four critical points (0, 0), (0, 8), (7, 0), and (4, 6). If x(t) and y(t) denote the number of rabbits and the number of squirrels, respectively, and if both populations are constant, it follows that the equations in (5) allow only three nontrivial possibilities: either no rabbits and 8 squirrels, or 7 rabbits and no squirrels, or 4 rabbits and 6 squirrels. In particular, the critical point (4, 6) describes the only possibility for the coexistence of constant nonzero populations of both species.

Phase Portraits

If the initial point (x0,y0) is not a critical point, then the corresponding trajectory is a curve in the xy-plane along which the point (x(t), y(t)) moves as t increases. It turns out that any trajectory not consisting of a single point is a nondegenerate curve with no self-intersections (Problem 30). We can exhibit qualitatively the behavior of solutions of the autonomous system in (1) by constructing a picture that shows its critical points together with a collection of typical solution curves or trajectories in the xy-plane. Such a picture is called a phase portrait (or phase plane picture) because it illustrates “phases” or xy-states of the system, and indicates how they change with time.

Another way of visualizing the system is to construct a slope field in the xy-phase plane by drawing typical line segments having slope

dydx=yx=G(x,y)F(x,y),

or a direction field by drawing typical vectors pointing the same direction at each point (x, y) as does the vector (F(x, y), G(x, y)). Such a vector field then indicates which direction along a trajectory to travel in order to “go with the flow” described by the system.

Remark

It is worth emphasizing that if our system of differential equations were not autonomous, then its critical points, trajectories, and direction vectors would generally be changing with time. In this event, the concrete visualization afforded by a (fixed) phase portrait or direction field would not be available to us. Indeed, this is a principal reason why an introductory study of nonlinear systems concentrates on autonomous ones.

Figure 9.1.1 shows a direction field and phase portrait for the rabbit-squirrel system of Example 1. The direction field arrows indicate the direction of motion of the point (x(t), y(t)). We see that, given any positive initial numbers x04 and y06 of rabbits and squirrels, this point moves along a trajectory approaching the critical point (4, 6) as t increases.

FIGURE 9.1.1.

Direction field and phase portrait for the rabbit-squirrel system x=14x2x2xy, y=16y2y2xy of Example 1.

Example 2

For the system

(7)x=xy,y=1x2

we see from the first equation that x=y and from the second that x=±1 at each critical point. Thus this system has the two critical points (1,1) and (1, 1). The direction field in Fig. 9.1.2 suggests that trajectories somehow “circulate” counterclockwise around the critical point (1,1), whereas it appears that some trajectories may approach, while others recede from, the critical point (1, 1). These observations are corroborated by the phase portrait in Fig. 9.1.3 for the system in (7).

FIGURE 9.1.2.

Direction field for the system in Eq. (7).

FIGURE 9.1.3.

Phase portrait for the system in Eq. (7).

Remark

One could carelessly write the critical points in Example 2 as (±1,±1) and then jump to the erroneous conclusion that the system in (7) has four rather than just two critical points. When feasible, a sure-fire way to determine the number of critical points of an autonomous system is to plot the curves F(x,y)=0 and G(x,y)=0 and then note their intersections, each of which represents a critical point of the system. For instance, Fig. 9.1.4 shows the curve (line) F(x,y)=xy=0 and the pair of lines x=+1 and x=1 that constitute the “curve” G(x,y)=1x2=0. The (only) two points of intersection (1,1) and (+1,+1) are then apparent.

FIGURE 9.1.4.

The two critical points (1,1) and (+1,+1) in Example 2 as the intersection of the curves F(x,y)=xy=0 and G(x,y)=1x2=0.

Critical Point Behavior

The behavior of the trajectories near an isolated critical point of an autonomous system is of particular interest. Figure 9.1.5 is a close-up view of Fig. 9.1.1 near the critical point (4, 6), and similarly Figs. 9.1.6 and 9.1.7 are close-ups of Fig. 9.1.3 near the critical points (1,1) and (1, 1), respectively. We notice immediately that although the systems underlying these phase portraits are nonlinear, each of these three magnifications bears a striking resemblance to one of the cases in our “gallery” Fig. 7.4.16 of phase plane portraits for linear constant-coefficient systems. Indeed, the three figures strongly resemble a nodal sink, a spiral source, and a saddle point, respectively.

FIGURE 9.1.5.

Close-up view of Fig. 9.1.1 near the critical point (4, 6).

FIGURE 9.1.6.

Close-up view of Fig. 9.1.3 near the critical point (1,1) .

FIGURE 9.1.7.

Close-up view of Fig. 9.1.3 near the critical point (1, 1).

These similarities are not a coincidence. Indeed, as we will explore in detail in the next section, the behavior of a nonlinear system near a critical point is generally similar to that of a corresponding linear constant-coefficient system near the origin. For this reason it is useful to extend the language of nodes, sinks, etc., introduced in Section 7.4 for linear constant-coefficient systems to the broader context of critical points of the two-dimensional system (1).

In general, the critical point (x,y) of the autonomous system in (1) is called a node provided that

  • Either every trajectory approaches (x,y) as t+ or every trajectory recedes from (x,y) as t+, and

  • Every trajectory is tangent at (x,y) to some straight line through the critical point.

As with linear constant-coefficient systems, a node is said to be proper provided that no two different pairs of “opposite” trajectories are tangent to the same straight line through the critical point. On the other hand, the critical point (4, 6) of the system in Eq. (5), shown in Figs. 9.1.1 and 9.1.5, is an improper node; as those figures suggest, virtually all of the trajectories approaching this critical point share a common tangent line at that point.

Likewise, a node is further called a sink if all trajectories approach the critical point, a source if all trajectories recede (or emanate) from it. Thus the critical point (4, 6) in Figs. 9.1.1 and 9.1.5 is a nodal sink, whereas the critical point (1,1) in Figs. 9.1.3 and 9.1.6 is a source (more specifically, a spiral source). The critical point (1, 1) in Figs. 9.1.3 and 9.1.7, on the other hand, is a saddle point.

Stability

A critical point (x,y) of the autonomous system in (1) is said to be stable provided that if the initial point (x0,y0) is sufficiently close to (x,y), then (x(t), y(t)) remains close to (x,y) for all t>0. In vector notation, with x(t)=(x(t),y(t)), the distance between the initial point x0=(x0,y0) and the critical point x=(x,y) is

|x0x|=(x0x)2+(y0y)2.

Thus the critical point x is stable provided that, for each ϵ>0, there exists δ>0 such that

(8)|x0x|=δimplies that|x(t)x|<ϵ

for all t>0. Note that the condition in (8) certainly holds if x(t)x as t+, as in the case of a nodal sink such as the critical point (4, 6) in Figs. 9.1.1 and 9.1.5. Thus this nodal sink can also be described as a stable node.

The critical point (x,y) is called unstable if it is not stable. The two critical points at (1,1) and (1, 1) shown in Figs. 9.1.3, 9.1.6, and 9.1.7 are both unstable, because, loosely speaking, in neither of these cases can we guarantee that a trajectory will remain near the critical point simply by requiring the trajectory to begin near the critical point.

If (x,y) is a critical point, then the equilibrium solution x(t)x, y(t)y is called stable or unstable depending on the nature of the critical point. In applications the stability of an equilibrium solution is often a crucial matter. For instance, suppose in Example 1 that x(t) and y(t) denote the rabbit and squirrel populations, respectively, in hundreds. We will see in Section 9.3 that the critical point (4, 6) in Fig. 9.1.1 is stable. It follows that if we begin with close to 400 rabbits and 600 squirrels—rather than exactly these equilibrium values—then for all future time there will remain close to 400 rabbits and close to 600 squirrels. Thus the practical consequence of stability is that slight changes (perhaps due to random births and deaths) in the equilibrium populations will not so upset the equilibrium as to result in large deviations from the equilibrium solutions.

It is possible for trajectories to remain near a stable critical point without approaching it, as Example 3 shows.

Example 3

Undamped mass-spring system Consider a mass m that oscillates without damping on a spring with Hooke’s constant k, so that its position function x(t) satisfies the differential equation x+ω2x=0 (where ω2=k/m). If we introduce the velocity y=dx/dt of the mass, we get the system

(9)dxdt=y,dydt=ω2x

with general solution

(10a)
(10b)
x(t)=Acosωt+Bsinωt,y(t)=Aωsinωt+Bωcosωt.

With C=A2+B2, A=Ccos α, and B=Csin α, we can rewrite the solution in (10) in the form

(11a)
(11b)
x(t)=Ccos(ωtα),y(t)=ωCsin(ωtα),

so it becomes clear that each trajectory other than the critical point (0, 0) is an ellipse with equation of the form

(12)x2C2+y2ω2C2=1.

As illustrated by the phase portrait in Fig. 9.1.8 (where ω=12), each point (x0,y0) other than the origin in the xy-plane lies on exactly one of these ellipses, and each solution (x(t), y(t)) traverses the ellipse through its initial point (x0,y0) in the clockwise direction with period P=2π/ω. (It is clear from (11) that x(t+P)=x(t) and y(t+P)=y(t) for all t.) Thus each nontrivial solution of the system in (9) is periodic and its trajectory is a simple closed curve enclosing the critical point at the origin.

FIGURE 9.1.8.

Direction field and elliptical trajectories for the system x=y, y=14x. The origin is a stable center.

Figure 9.1.9 shows a typical elliptical trajectory in Example 3, with its minor semiaxis denoted by δ and its major semiaxis by ϵ. We see that if the initial point (x0,y0) lies within distance δ of the origin—so that its elliptical trajectory lies inside the one shown—then the point (x(t),y(t)) always remains within distance ϵ of the origin. Hence the origin (0, 0) is a stable critical point of the system x=y, y=ω2x, despite the fact that no single trajectory approaches the point (0, 0). A stable critical point surrounded by simple closed trajectories representing periodic solutions is called a (stable) center.

FIGURE 9.1.9.

If the initial point (x0,y0) lies within distance δ of the origin, then the point (x(t), y(t)) stays within distance ϵ of the origin.

Asymptotic Stability

The critical point (x,y) is called asymptotically stable if it is stable and, moreover, every trajectory that begins sufficiently close to (x,y) also approaches (x,y) as t+. That is, there exists δ>0 such that

(13)|xx|<δimplies thatlimtx(t)=x,

where x0=(x0,y0), x=(x,y), and x(t)=(x(t),y(t)) is a solution with x(0)=x0.

Remark

The stable node shown in Figs. 9.1.1 and 9.1.5 is asymptotically stable because every trajectory approaches the critical point (4, 6) as t+. The center (0, 0) shown in Fig. 9.1.8 is stable but not asymptotically stable, because however small an elliptical trajectory we consider, a point moving around this ellipse does not approach the origin. Thus asymptotic stability is a stronger condition than mere stability.

Now suppose that x(t) and y(t) denote coexisting populations for which (x,y) is an asymptotically stable critical point. Then if the initial populations x0 and y0 are sufficiently close to x and y, respectively, it follows that both

(14)limtx(t)=xandlimty(t)=y.

That is, x(t) and y(t) actually approach the equilibrium populations x and y as t+, rather than merely remaining close to those values.

For a mechanical system as in Example 3, a critical point represents an equilibrium state of the system—if the velocity y=x and the acceleration y=x vanish simultaneously, then the mass remains at rest with no net force acting on it. Stability of a critical point concerns the question whether, when the mass is displaced slightly from its equilibrium, it

  1. Moves back toward the equilibrium point as t+,

  2. Merely remains near the equilibrium point without approaching it, or

  3. Moves farther away from equilibrium.

In Case 1 the critical [equilibrium] point is asymptotically stable; in Case 2 it is stable but not asymptotically so; in Case 3 it is an unstable critical point. A marble balanced on the top of a soccer ball is an example of an unstable critical point. A mass on a spring with damping illustrates the case of asymptotic stability of a mechanical system. The mass-and-spring without damping in Example 3 is an example of a system that is stable but not asymptotically stable.

Example 4

Damped mass-spring system Suppose that m=1 and k=2 for the mass and spring of Example 3 and that the mass is attached also to a dashpot with damping constant c=2. Then its displacement function x(t) satisfies the second-order equation

(15)x(t)+2x(t)+2x(t)=0.

With y=x we obtain the equivalent first-order system

(16)dxdt=y,dydt=2x2y

with critical point (0, 0). The characteristic equation r2+2r+2=0 of Eq. (15) has roots 1+i and 1i, so the general solution of the system in (16) is given by

(17a)
(17b)
x(t)=et(Acost+Bsint)=Cetcos(tα),y(t)=et[(BA)cost(A+B)sint]=C2etsin(tα+14π),

where C=A2+B2 and α=tan1(B/A). We see that x(t) and y(t) oscillate between positive and negative values and that both approach zero as t+. Thus a typical trajectory spirals inward toward the origin, as illustrated by the spiral in Fig. 9.1.10.

FIGURE 9.1.10.

A stable spiral point and one nearby trajectory.

It is clear from (17) that the point (x(t), y(t)) approaches the origin as t+, so it follows that (0, 0) is an asymptotically stable critical point for the system x=y, y=2x2y of Example 4. Such an asymptotically stable critical point—around which the trajectories spiral as they approach it—is called a stable spiral point (or a spiral sink). In the case of a mass–spring—dashpot system, a spiral sink is the manifestation in the phase plane of the damped oscillations that occur because of resistance.

If the arrows in Fig. 9.1.10 were reversed, we would see a trajectory spiraling outward from the origin. An unstable critical point—around which the trajectories spiral as they emanate and recede from it—is called an unstable spiral point (or a spiral source). Example 5 shows that it also is possible for a trajectory to spiral into a closed trajectory—a simple closed solution curve that represents a periodic solution (like the elliptical trajectories in Fig. 9.1.8).

Example 5

Consider the system

(18)dxdt=ky+x(1x2y2),dydt=kx+y(1x2y2).

In Problem 21 we ask you to show that (0, 0) is its only critical point. This system can be solved explicitly by introducing polar coordinates x=rcos θ, y=rsin θ, as follows. First note that

dθdt=ddt(arctanyx)=xyxyx2+y2.

Then substitute the expressions given in (18) for x and y to obtain

dθdt=k(x2+y2)x2+y2=k.

It follows that

(19)θ(t)=kt+θ0,where θ0=θ(0).

Then differentiation of r2=x2+y2 yields

2rdrdt=2xdxdt+2ydydt=2(x2+y2)(1x2y2)=2r2(1r2),

so r=r(t) satisfies the differential equation

(20)drdt=r(1r2).

In Problem 22 we ask you to derive the solution

(21)r(t)=r0r02+(1r02)e2t,

where r0=r(0). Thus the typical solution of Eq. (18) may be expressed in the form

(22)x(t)=r(t)cos(kt+θ0),y(t)=r(t)sin(kt+θ0).

If r0=1, then Eq. (21) gives r(t)1 (the unit circle). Otherwise, if r0>0, then Eq. (21) implies that r(t)1 as t+. Hence the trajectory defined in (22) spirals in toward the unit circle if r0>1 and spirals out toward this closed trajectory if 0<r0<1. Figure 9.1.11 shows a trajectory spiraling outward from the origin and four trajectories spiraling inward, all approaching the closed trajectory r(t)1.

FIGURE 9.1.11.

Spiral trajectories of the system in Eq. (18) with k=5.

Under rather general hypotheses it can be shown that there are four possibilities for a nondegenerate trajectory of the autonomous system

dxdt=F(x,y),dydt=G(x,y).

The four possibilities are these:

  1. (x(t), y(t)) approaches a critical point as t+.

  2. (x(t), y(t)) is unbounded with increasing t.

  3. (x(t), y(t)) is a periodic solution with a closed trajectory.

  4. (x(t), y(t)) spirals toward a closed trajectory as t+.

As a consequence, the qualitative nature of the phase plane picture of the trajectories of an autonomous system is determined largely by the locations of its critical points and by the behavior of its trajectories near its critical points. We will see in Section 9.2 that, subject to mild restrictions on the functions F and G, each isolated critical point of the system x=F(x,y), y=G(x,y) resembles qualitatively one of the examples of this section—it is either a node (proper or improper), a saddle point, a center, or a spiral point.

9.1 Problems

In Problems 1 through 8, find the critical point or points of the given autonomous system, and thereby match each system with its phase portrait among Figs. 9.1.12 through 9.1.19.

  1. dxdt=2xy,dydt=x3y

  2. dxdt=xy,dydt=x+3y4

  3. dxdt=x2y+3,dydt=xy+2

  4. dxdt=2x2y4,dydt=x+4y+3

  5. dxdt=1y2,dydt=x+2y

  6. dxdt=24x15y,dydt=4x2

FIGURE 9.1.12.

Spiral point (2,1) and saddle point (2,1).

FIGURE 9.1.13.

Spiral point (1,1).

FIGURE 9.1.14.

Saddle point (0, 0).

FIGURE 9.1.15.

Spiral point (0, 0); saddle points (2,1) and (2, 1).

FIGURE 9.1.16.

Node (1, 1).

FIGURE 9.1.17.

Spiral point (1,1), saddle point (0, 0), and node (1,1).

FIGURE 9.1.18.

Spiral point (2,23) and saddle point (2,25).

FIGURE 9.1.19.

Stable center (1,1).

  1. dxdt=x2y,dydt=4xx3

  2. dxdt=xyx2+xy,dydt=yx2

In Problems 9 through 12, find each equilibrium solution x(t)x0 of the given second-order differential equation x+f(x,x)=0. Use a computer system or graphing calculator to construct a phase portrait and direction field for the equivalent first-order system x=y, y=f(x,y). Thereby ascertain whether the critical point (x0,0) looks like a center, a saddle point, or a spiral point of this system.

  1. x+4xx3=0

  2. x+2x+x+4x3=0

  3. x+3x+4sin x=0

  4. x+(x21)x+x=0

Solve each of the linear systems in Problems 13 through 20 to determine whether the critical point (0, 0) is stable, asymptotically stable, or unstable. Use a computer system or graphing calculator to construct a phase portrait and direction field for the given system. Thereby ascertain the stability or instability of each critical point, and identify it visually as a node, a saddle point, a center, or a spiral point.

  1. dxdt=2x,dydt=2y

  2. dxdt=2x,dydt=2y

  3. dxdt=2x,dydt=y

  4. dxdt=x,dydt=3y

  5. dxdt=y,dydt=x

  6. dxdt=y,dydt=4x

  7. dxdt=2y,dydt=2x

  8. dxdt=y,dydt=5x4y

  9. Verify that (0, 0) is the only critical point of the system in Example 6.

  10. Separate variables in Eq. (20) to derive the solution in (21).

In Problems 23 through 26, a system dx/dt=F(x,y), dy/dt=G(x,y) is given. Solve the equation

dydx=G(x,y)F(x,y)

to find the trajectories of the given system. Use a computer system or graphing calculator to construct a phase portrait and direction field for the system, and thereby identify visually the apparent character and stability of the critical point (0, 0) of the given system.

  1. dxdt=y,dydt=x

  2. dxdt=y(1+x2+y2),dydt=x(1+x2+y2)

  3. dxdt=4y(1+x2+y2),dydt=x(1+x2+y2)

  4. dxdt=y3ex+y,dydt=x3ex+y

  5. Let (x(t), y(t)) be a nontrivial solution of the nonautonomous system

    dxdt=y,dydt=tx.

    Suppose that ϕ(t)=x(t+γ) and ψ(t)=y(t+γ), where γ0. Show that (ϕ(t),ψ(t)) is not a solution of the system.

Problems 28 through 30 deal with the system

dxdt=F(x,y),dydt=G(x,y)

in a region where the functions F and G are continuously differentiable, so for each number a and point (x0, y0), there is a unique solution with x(a)=x0 and y(a)=y0.

  1. Suppose that (x(t), y(t)) is a solution of the autonomous system and that γ0. Define ϕ(t)=x(t+γ) and ψ(t)=y(t+γ). Then show (in contrast with the situation in Problem 27) that (ϕ(t),ψ(t)) is also a solution of the system. Thus autonomous systems have the simple but important property that a “t-translate” of a solution is again a solution.

  2. Let (x1(t),y1(t)) and (x2(t),y2(t)) be two solutions having trajectories that meet at the point (x0,y0); thus x1(a)=x2(b)=x0 and y1(a)=y2(b)=y0 for some values a and b of t. Define

    x3(t)=x2(t+γ)andy3(t)=y2(t+γ),

    where γ=ba, so (x2(t),y2(t)) and (x3(t),y3(t)) have the same trajectory. Apply the uniqueness theorem to show that (x1(t),y1(t)) and (x3(t),y3(t)) are identical solutions. Hence the original two trajectories are identical. Thus no two different trajectories of an autonomous system can intersect.

  3. Suppose that the solution (x1(t),y1(t)) is defined for all t and that its trajectory has an apparent self-intersection:

    x1(a)=x1(a+P)=x0,y1(a)=y1(a+P)=y0

    for some P>0. Introduce the solution

    x2(t)=x1(t+P),y2(t)=y1(t+P),

    and then apply the uniqueness theorem to show that

    x1(t+P)=x1(t)andy1(t)=y1(t+P)

    for all t. Thus the solution (x1(t),y1(t)) is periodic with period P and has a closed trajectory. Consequently a solution of an autonomous system either is periodic with a closed trajectory, or else its trajectory never passes through the same point twice.

9.1 Application Phase Plane Portraits and First-Order Equations

Consider a first-order differential equation of the form

(1)dydx=G(x,y)F(x,y),

which may be difficult or impossible to solve explicitly. Its solution curves can nevertheless be plotted as trajectories of the corresponding autonomous two-dimen-sional system

(2)dxdt=F(x,y),dydt=G(x,y).

Most ODE plotters can routinely generate phase portraits for autonomous systems. Those appearing in this chapter were plotted using programs that are free for educational use. For instance, the Matlab program pplane illustrated in Fig. 9.1.20 can be found at math.rice.edu/~dfield. Another freely available and user-friendly Matlab-based ODE package with similar graphical capabilities is Iode (www.math.uiuc.edu/iode).

FIGURE 9.1.20.

Matlab pplane menu entries to plot a direction field and phase portrait for the system x=y,y=14x (as shown in Fig. 9.1.8).

For example, to plot solution curves for the differential equation

(3)dydx=2xyy2x22xy,

we plot trajectories of the system

(4)dxdt=x22xy,dydt=2xyy2.

The result is shown in Fig. 9.1.21.

FIGURE 9.1.21.

Phase portrait for the system in Eq. (4).

Plot similarly some solution curves for the following differential equations.

  1. dydx=4x5y2x+3y

  2. dydx=4x5y2x3y

  3. dydx=4x3y2x5y

  4. dydx=2xyx2y2

  5. dydx=x2+2xyy2+2xy

Now construct some examples of your own. Homogeneous functions like those in Problems 1 through 5—rational functions with numerator and denominator of the same degree in x and y—work well. The differential equation

(5)dydx=25x+y(1x2y2)(4x2y2)25y+x(1x2y2)(4x2y2)

of this form generalizes Example 5 in this section but would be inconvenient to solve explicitly. Its phase portrait (Fig. 9.1.22) shows two periodic closed trajectories—the circles r=1 and r=2. Anyone want to try for three circles?

FIGURE 9.1.22.

Phase portrait for the system corresponding to Eq. (5).

9.2 Linear and Almost Linear Systems

We now discuss the behavior of solutions of the autonomous system

(1)dxdt=f(x,y),dydt=g(x,y)

near an isolated critical point (x0,y0) where f(x0,y0)=g(x0,y0)=0. A critical point is called isolated if some neighborhood of it contains no other critical point. We assume throughout that the functions f and g are continuously differentiable in a neighborhood of (x0,y0).

We can assume without loss of generality that x0=y0=0. Otherwise, we make the substitutions u=xx0, v=yy0. Then dx/dt=du/dt and dy/dt=dv/dt, so (1) is equivalent to the system

(2)dudt=f(u+x0,v+y0)=f1(u,v),dvdt=g(u+x0,v+y0)=g1(u,v)

that has (0, 0) as an isolated critical point.

Example 1

The system

(3)dxdt=3xx2xy=x(3xy),dydt=y+y23xy=y(13x+y)

has (1, 2) as one of its critical points. We substitute u=x1, v=y2; that is, x=u+1, y=v+2. Then

3xy=3(u+1)(v+2)=uv

and

13x+y=13(u+1)+(v+2)=3u+v,

so the system in (3) takes the form

(4)dudt=(u+1)(uv)=uvu2uv,dvdt=(v+2)(3u+v)=6u+2v+v23uv

and has (0, 0) as a critical point. If we can determine the trajectories of the system in (4) near (0, 0), then their translations under the rigid motion that carries (0, 0) to (1, 2) will be the trajectories near (1, 2) of the original system in (3). This equivalence is illustrated by Fig. 9.2.1 (which shows computer-plotted trajectories of the system in (3) near the critical point (1, 2) in the xy-plane) and Fig. 9.2.2 (which shows computer-plotted trajectories of the system in (4) near the critical point (0, 0) in the uv-plane).

FIGURE 9.2.1.

The saddle point (1, 2) for the system x=3xx2xy,y=y+y23xy of Example 1.

FIGURE 9.2.2.

The saddle point (0, 0) for the equivalent system u=uvu2uv,v=6u+2v+v23uv.

Figures 9.2.1 and 9.2.2 illustrate the fact that the solution curves of the xy-system in (1) are simply the images under the translation (u,v)(u+x0,v+y0) of the solution curves of the uv-system in (2). Near the two corresponding critical points—(x0,y0) in the xy-plane and (0, 0) in the uv-plane—the two phase portraits therefore look precisely the same.

Linearization Near a Critical Point

Taylor’s formula for functions of two variables implies that—if the function f(x, y) is continuously differentiable near the fixed point (x0,y0)—then

f(x0+u,y0+v)=f(x0,y0)+fx(x0,y0)u+fy(x0,y0)v+r(u,v)

where the “remainder term” r(u, v) satisfies the condition

lim(u,v)(0,0)r(u,v)u2+v2=0.

(Note that this condition would not be satisfied if r(u, v) were a sum containing either constants or terms linear in u or v. In this sense, r(u, v) consists of the “nonlinear part” of the function f(x0+u,y0+v) of u and v.)

If we apply Taylor’s formula to both f and g in (2) and assume that (x0,y0) is an isolated critical point so f(x0,y0)=g(x0,y0)=0, the result is

(5)dudt=fx(x0,y0)u+fy(x0,y0)v+r(u,v),dvdt=gx(x0,y0)u+gy(x0,y0)v+s(u,v)

where r(u, v) and the analogous remainder term s(u, v) for g satisfy the condition

(6)lim(u,v)(0,0)r(u,v)u2+v2=lim(u,v)(0,0)s(u,v)u2+v2=0.

Then, when the values u and v are small, the remainder terms r(u, v) and s(u, v) are very small (being small even in comparison with u and v).

If we drop the presumably small nonlinear terms r(u, v) and s(u, v) in (5), the result is the linear system

(7)dudt=fx(x0,y0)u+fy(x0,y0)v,dvdt=gx(x0,y0)u+gy(x0,y0)v

whose constant coefficients (of the variables u and v) are the values fx(x0,y0), fy(x0,y0) and gx(x0,y0), gy(x0,y0) of the functions f and g at the critical point (x0,y0). Because (5) is equivalent to the original (and generally) nonlinear system u=f(x0+u,y0+v), v=g(x0+u,y0+v) in (2), the conditions in (6) suggest that the linearized system in (7) closely approximates the given nonlinear system when (u, v) is close to (0, 0).

Assuming that (0, 0) is also an isolated critical point of the linear system, and that the remainder terms in (5) satisfy the condition in (6), the original system x=f(x,y), y=g(x,y) is said to be almost linear at the isolated critical point (x0,y0). In this case, its linearization at (x0,y0) is the linear system in (7). In short, this linearization is the linear system u=Ju (where u=[uv]T) whose coefficient matrix is the so-called Jacobian matrix

(8)J(x0,y0)=[fx(x0,y0)fy(x0,y0)gx(x0,y0)gy(x0,y0)]

of the functions f and g, evaluated at the point (x0,y0).

Example 1

Continued In (3) we have f(x,y)=3xx2xy and g(x,y)=y+y23xy. Then

J(x,y)=[32xyx3y1+2y3x],soJ(1,2)=[1162].

Hence the linearization of the system x=3xx2xy, y=y+y23xy at its critical point (1, 2) is the linear system

u=uv,v=6u+2v

that we get when we drop the nonlinear (quadratic) terms in (4).

It turns out that in most (though not all) cases, the phase portrait of an almost linear system near an isolated critical point (x0,y0) strongly resembles—qualitatively—the phase portrait of its linearization near the origin. Consequently, the first step toward understanding general autonomous systems is to characterize the isolated critical points of linear systems.

Isolated Critical Points of Linear Systems

In Section 7.4 we used the eigenvalue-eigenvector method to study the 2×2 linear system

(9)[xy]=[abcd][xy]

with constant-coefficient matrix A. The origin (0, 0) is a critical point of the system regardless of the matrix A, but if we further require the origin to be an isolated critical point, then (by a standard theorem of linear algebra) the determinant adbc of A must be nonzero. From this we can conclude that the eigenvalues λ1 and λ2 of A must be nonzero. Indeed, λ1 and λ2 are the solutions of the characteristic equation

(10)det(AλI)=|aλbcdλ|=(aλ)(dλ)bc=λ2(a+d)λ+(adbc)=0,

and the fact that adbc0 implies that λ=0 cannot satisfy Eq. (10); hence λ1 and λ2 are nonzero. The converse also holds: If the characteristic equation (10) has no zero solution—that is, if all eigenvalues of the matrix A are nonzero—then the determinant adbc is nonzero. Altogether, we conclude that the origin (0, 0) is an isolated critical point of the system in Eq. (9) if and only if the eigenvalues of A are all nonzero. Our study of this critical point can be divided, therefore, into the five cases listed in the table in Fig. 9.2.3. This table also gives the type of each critical point as found in Section 7.4 and shown in our gallery Fig. 7.4.16 of typical phase plane portraits:

FIGURE 9.2.3.

Classification of the isolated critical point (0, 0) of the two-dimensional system x=Ax.

Eigenvalues of A Type of Critical Point
Real, unequal, same sign Improper node
Real, unequal, opposite sign Saddle point
Real and equal Proper or improper node
Complex conjugate Spiral point
Pure imaginary Center

Closer inspection of that gallery, however, reveals a striking connection between the stability properties of the critical point and the eigenvalues λ1 and λ2 of A. For example, if λ1 and λ2 are real, unequal, and negative, then the origin represents an improper nodal sink; because all trajectories approach the origin as t+, the critical point is asymptotically stable. Likewise, if λ1 and λ2 are real, equal, and negative, then the origin is a proper nodal sink, and is again asymptotically stable. Further, if λ1 and λ2 are complex conjugate with negative real part, then the origin is a spiral sink, and is once more asymptotically stable. All three of these cases can be captured as follows: If the real parts of λ1 and λ2 are negative, then the origin is an asymptotically stable critical point. (Of course, if λ1 and λ2 are real, then they are themselves their real parts.)

Similar generalizations can be made for other combinations of signs of the real parts of λ1 and λ2. Indeed, as the table in Fig. 9.2.4 shows, the stability properties of the isolated critical point (0, 0) of the system in Eq. (9) are always determined by the signs of the real parts of λ1 and λ2. (We invite you to use the gallery in Fig. 7.4.16 to verify the conclusions in the table.)

FIGURE 9.2.4.

Stability properties of the isolated critical point (0, 0) of the system in Eq. (9) with nonzero eigenvalues λ1 and λ2.

Real Parts of λ1 and λ2 Type of Critical Point Stability
Both negative
  • Proper or improper nodal sink, or

  • Spiral sink

Asymptotically stable
Both zero (i.e., λ1 and λ2 are given by ±iq with q0)
  • Center

Stable but not asymptotically stable
At least one positive
  • Proper or improper nodal source, or

  • Spiral source, or

  • Saddle point

Unstable

These findings are summarized in Theorem 1:

It is worthwhile to consider the effect of small perturbations in the coefficients a, b, c, and d of the linear system in (11), which result in small perturbations of the eigenvalues λ1 and λ2. If these perturbations are sufficiently small, then positive real parts (of λ1 and λ2) remain positive and negative real parts remain negative. Hence an asymptotically stable critical point remains asymptotically stable and an unstable critical point remains unstable. Part 2 of Theorem 1 is therefore the only case in which arbitrarily small perturbations can affect the stability of the critical point (0, 0). In this case pure imaginary roots λ1, λ2=±qi of the characteristic equation can be changed to nearby complex roots μ1, μ2=r±si, with r either positive or negative (see Fig. 9.2.5). Consequently, a small perturbation of the coefficients of the linear system in (11) can change a stable center to a spiral point that is either unstable or asymptotically stable.

FIGURE 9.2.5.

The effects of perturbation of pure imaginary roots.

There is one other exceptional case in which the type, though not the stability, of the critical point (0, 0) can be altered by a small perturbation of its coefficients. This is the case with λ1=λ2, equal roots that (under a small perturbation of the coefficients) can split into two roots μ1 and μ2, which are either complex conjugates or unequal real roots (see Fig. 9.2.6). In either case, the sign of the real parts of the roots is preserved, so the stability of the critical point is unaltered. Its nature may change, however; the table in Fig. 9.2.3 shows that a node with λ1=λ2 can either remain a node (if μ1 and μ2 are real) or change to a spiral point (if μ1 and μ2 are complex conjugates).

FIGURE 9.2.6.

The effects of perturbation of real equal roots.

Suppose that the linear system in (11) is used to model a physical situation. It is unlikely that the coefficients in (11) can be measured with total accuracy, so let the unknown precise linear model be

(11*)dxdt=ax+by,dydt=cx+dy.

If the coefficients in (11) are sufficiently close to those in (11), it then follows from the discussion in the preceding paragraph that the origin (0, 0) is an asymptotically stable critical point for (11) if it is an asymptotically stable critical point for (11), and is an unstable critical point for (11) if it is an unstable critical point for (11). Thus in this case the approximate model in (11) and the precise model in (11) predict the same qualitative behavior (with respect to asymptotic stability versus instability).

Almost Linear Systems

Recall that we first encountered an almost linear system at the beginning of this section, when we used Taylor’s formula to write the nonlinear system (2) in the almost linear form (5) which led to the linearization (7) of the original nonlinear system. In case the nonlinear system x=f(x,y), y=g(x,y) has (0, 0) as an isolated critical point, the corresponding almost linear system is

(12)dxdt=ax+by+r(x,y),dydt=cx+dy+s(x,y),

where a=fx(0,0), b=fy(0,0) and c=gx(0,0), d=gy(0,0); we assume also that adbc0. Theorem 2, which we state without proof, essentially implies that—with regard to the type and stability of the critical point (0, 0)—the effect of the small nonlinear terms r(x, y) and s(x, y) is equivalent to the effect of a small perturbation in the coefficients of the associated linear system in (11).

Thus, if λ1λ2 and Re(λ1)0, then the type and stability of the critical point of the almost linear system in (12) can be determined by analysis of its associated linear system in (11), and only in the case of pure imaginary eigenvalues is the stability of (0, 0) not determined by the linear system. Except in the sensitive cases λ1=λ2 and Re(λi)=0, the trajectories near (0, 0) will resemble qualitatively those of the associated linear system—they enter or leave the critical point in the same way, but may be “deformed” in a nonlinear manner. The table in Fig. 9.2.7 summarizes the situation.

FIGURE 9.2.7.

Classification of critical points of an almost linear system.

Eigenvalues λ1, λ2 for the Linearized System Type of Critical Point of the Almost Linear System
λ1<λ2<0 Stable improper node
λ1=λ2<0 Stable node or spiral point
λ1<0<λ2 Unstable saddle point
λ1=λ2>0 Unstable node or spiral point
λ1>λ2>0 Unstable improper node
λ1,λ2=a±bi(a<0) Stable spiral point
λ1,λ2=a±bi(a>0) Unstable spiral point
λ1,λ2=±bi Stable or unstable, center or spiral point

An important consequence of the classification of cases in Theorem 2 is that a critical point of an almost linear system is asymptotically stable if it is an asymptotically stable critical point of the linearization of the system. Moreover, a critical point of the almost linear system is unstable if it is an unstable critical point of the linearized system. If an almost linear system is used to model a physical situation, then—apart from the sensitive cases mentioned earlier—it follows that the qualitative behavior of the system near a critical point can be determined by examining its linearization.

Example 2

Determine the type and stability of the critical point (0, 0) of the almost linear system

(13)dxdt=4x+2y+2x23y2,dydt=4x3y+7xy.

Solution

The characteristic equation for the associated linear system (obtained simply by deleting the quadratic terms in (13)) is

(4λ)(3λ)8=(λ5)(λ+4)=0,

so the eigenvalues λ1=5 and λ2=4 are real, unequal, and have opposite signs. By our discussion of this case we know that (0, 0) is an unstable saddle point of the linear system, and hence by Part 3 of Theorem 2, it is also an unstable saddle point of the almost linear system in (13). The trajectories of the linear system near (0, 0) are shown in Fig. 9.2.8, and those of the nonlinear system in (13) are shown in Fig. 9.2.9. Figure 9.2.10 shows a phase portrait of the nonlinear system in (13) from a “wider view.” In addition to the saddle point at (0, 0), there are spiral points near the points (0.279, 1.065) and (0.933,1.057), and a node near (2.354,0.483).

FIGURE 9.2.8.

Trajectories of the linearized system of Example 2.

FIGURE 9.2.9.

Trajectories of the original almost linear system of Example 2.

FIGURE 9.2.10.

Phase portrait for the almost linear system in Eq. (13).

We have seen that the system x=f(x,y), y=g(x,y) with isolated critical point (x0,y0) transforms via the substitution x=u+x0, y=v+y0 to an equivalent uv-system with corresponding critical point (0, 0) and linearization u=Ju, whose coefficient matrix J is the Jacobian matrix in (8) of the functions f and g at (x0,y0). Consequently we need not carry out the substitution explicitly; instead, we can proceed directly to calculate the eigenvalues of J preparatory to application of Theorem 2.

Example 3

Determine the type and stability of the critical point (4, 3) of the almost linear system

(14)dxdt=3310x3y+x2,dydt=18+6x+2yxy.

Solution

With f(x,y)=3310x3y+x2, g(x,y)=18+6x+2yxy and x0=4, y0=3 we have

J(x,y)=[10+2x36y2x],soJ(4,3)=[2332].

The associated linear system

(15)dudt=2u3v,dvdt=3u2v

has characteristic equation (λ+2)2+9=0, with complex conjugate roots λ=2±3i. Hence (0, 0) is an asymptotically stable spiral point of the linear system in (15), so Theorem 2 implies that (4, 3) is an asymptotically stable spiral point of the original almost linear system in (14). Figure 9.2.11 shows some typical trajectories of the linear system in (15), and Fig. 9.2.12 shows how this spiral point fits into the phase portrait for the original almost linear system in (14).

FIGURE 9.2.11.

Spiral trajectories of the linear system in Eq. (15).

FIGURE 9.2.12.

Phase portrait for the almost linear system in Eq. (14).

9.2 Problems

In Problems 1 through 10, apply Theorem 1 to determine the type of the critical point (0, 0) and whether it is asymptotically stable, stable, or unstable. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given linear system.

  1. dxdt=2x+y,dydt=x2y

  2. dxdt=4xy,dydt=2x+y

  3. dxdt=x+2y,dydt=2x+y

  4. dxdt=3x+y,dydt=5xy

  5. dxdt=x2y,dydt=2x3y

  6. dxdt=5x3y,dydt=3xy

  7. dxdt=3x2y,dydt=4xy

  8. dxdt=x3y,dydt=6x5y

  9. dxdt=2x2y,dydt=4x2y

  10. dxdt=x2y,dydt=5xy

Each of the systems in Problems 11 through 18 has a single critical point (x0,y0). Apply Theorem 2 to classify this critical point as to type and stability. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given system.

  1. dxdt=x2y,dydt=3x4y2

  2. dxdt=x2y8,dydt=x+4y+10

  3. dxdt=2xy2,dydt=3x2y2

  4. dxdt=x+y7,dydt=3xy5

  5. dxdt=xy,dydt=5x3y2

  6. dxdt=x2y+1,dydt=x+3y9

  7. dxdt=x5y5,dydt=xy3

  8. dxdt=4x5y+3,dydt=5x4y+6

In Problems 19 through 28, investigate the type of the critical point (0, 0) of the given almost linear system. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait. Also, describe the approximate locations and apparent types of any other critical points that are visible in your figure. Feel free to investigate these additional critical points; you can use the computational methods discussed in the application material for this section.

  1. dxdt=x3y+2xy,dydt=4x6yxy

  2. dxdt=6x5y+x2,dydt=2xy+y2

  3. dxdt=x+2y+x2+y2,dydt=2x2y3xy

  4. dxdt=x+4yxy2,dydt=2xy+x2y

  5. dxdt=2x5y+x3,dydt=4x6y+y4

  6. dxdt=5x3y+y(x2+y2),dydt=5x+y(x2+y2)

  7. dxdt=x2y+3xy,dydt=2x3yx2y2

  8. dxdt=3x2yx2y2,dydt=2xy3xy

  9. dxdt=xy+x4y2,dydt=2xy+y4x2

  10. dxdt=3xy+x3+y3,dydt=13x3y+3xy

In Problems 29 through 32, find all critical points of the given system, and investigate the type and stability of each. Verify your conclusions by means of a phase portrait constructed using a computer system or graphing calculator.

  1. dxdt=xy,dydt=x2y

  2. dxdt=y1,dydt=x2y

  3. dxdt=y21,dydt=x3y

  4. dxdt=xy2,dydt=x2y

Bifurcations

The term bifurcation generally refers to something “splitting apart.” With regard to differential equations or systems involving a parameter, it refers to abrupt changes in the character of the solutions as the parameter is changed continuously. Problems 33 through 36 illustrate sensitive cases in which small perturbations in the coefficients of a linear or almost linear system can change the type or stability (or both) of a critical point.

  1. Consider the linear system

    dxdt=ϵxy,dydt=x+ϵy.

    Show that the critical point (0, 0) is (a) a stable spiral point if ϵ<0; (b) a center if ϵ=0; (c) an unstable spiral point if ϵ>0; Thus small perturbations of the system x=y,y=x can change both the type and stability of the critical point. Figures 9.2.13(a)(e) illustrate the loss of stability that occurs at ϵ=0 as the parameter increases from ϵ<0 to ϵ>0.

  2. Consider the linear system

    dxdt=x+ϵy,dydt=xy.

    Show that the critical point (0, 0) is (a) a stable spiral point if ϵ<0; (b) a stable node if 0ϵ<1. Thus small perturbations of the system x=x, y=xy can change the type of the critical point (0, 0) without changing its stability.

FIGURE 9.2.13(a).

Stable spiral with ϵ=0.2.

FIGURE 9.2.13(b).

Stable spiral with ϵ=0.05.

FIGURE 9.2.13(c).

Stable center with ϵ=0.

FIGURE 9.2.13(d).

Unstable spiral with ϵ=0.05.

FIGURE 9.2.13(e).

Unstable spiral with ϵ=0.2.

  1. This problem deals with the almost linear system

    dxdt=y+hx(x2+y2),dydt=x+hy(x2+y2),

    in illustration of the sensitive case of Theorem 2, in which the theorem provides no information about the stability of the critical point (0, 0). (a) Show that (0, 0) is a center of the linear system obtained by setting h=0. (b) Suppose that h0. Let r2=x2+y2, then apply the fact that

    xdxdt+ydydt=rdrdt

    to show that dr/dt=hr3. (c) Suppose that h=1. Integrate the differential equation in (b); then show that r0 as t+. Thus (0, 0) is an asymptotically stable critical point of the almost linear system in this case. (d) Suppose that h=+1. Show that r+ as t increases, so (0, 0) is an unstable critical point in this case.

  2. This problem presents the famous Hopf bifurcation for the almost linear system

    dxdt=ϵx+yx(x2+y2),dydt=x+ϵyy(x2+y2),

    which has imaginary characteristic roots λ=±i if ϵ=0. (a) Change to polar coordinates as in Example 5 of Section 9.1 to obtain the system r=r(ϵr2), θ=1. (b) Separate variables and integrate directly to show that if ϵ0, then r(t)0 as t+, so in this case the origin is a stable spiral point. (c) Show similarly that if ϵ>0, then r(t)ϵ as t+, so in this case the origin is an unstable spiral point. The circle r(t)ϵ itself is a closed periodic solution or limit cycle. Thus a limit cycle of increasing size is spawned as the parameter ϵ increases through the critical value 0.

  3. In the case of a two-dimensional system that is not almost linear, the trajectories near an isolated critical point can exhibit a considerably more complicated structure than those near the nodes, centers, saddle points, and spiral points discussed in this section. For example, consider the system

    (16)dxdt=x(x32y3),dydt=y(2x3y3)

    having (0, 0) as an isolated critical point. This system is not almost linear because (0, 0) is not an isolated critical point of the trivial associated linear system x=0, y=0. Solve the homogeneous first-order equation

    dydx=y(2x3y3)x(x32y3)

    to show that the trajectories of the system in (16) are folia of Descartes of the form

    x3+y3=3cxy,

    where c is an arbitrary constant (Fig. 9.2.14).

    FIGURE 9.2.14.

    Trajectories of the system in Eq. (16).

  4. First note that the characteristic equation of the 2×2 matrix A can be written in the form λ2Tλ+D=0, where D is the determinant of A and the trace T of the matrix A is the sum of its two diagonal elements. Then apply Theorem 1 to show that the type of the critical point (0, 0) of the system x=Ax is determined-as indicated in Fig. 9.2.15 -by the location of the point (T,D) in the trace-determinant plane with horizontal T-axis and vertical D-axis.

    FIGURE 9.2.15.

    The critical point (0, 0) of the system x=Ax is a

    • spiral sink or source if the point (T, D) lies above the parabola T2=4D but off the D-axis;

    • stable center if (T, D) lies on the positive D-axis;

    • nodal sink or source if (T, D) lies between the parabola and the T-axis;

    • saddle point if (T, D) lies beneath the T-axis.

9.2 Application Phase Plane Portraits of Almost Linear Systems

Interesting and complicated phase portraits often result from simple nonlinear perturbations of linear systems. For instance, Fig. 9.2.16 shows a phase portrait for the almost linear system

(1)dxdt=ycos(x+y1),dydt=xcos(xy+1).

Among the seven critical points marked with dots, we see

FIGURE 9.2.16.

Phase plane portrait for the system in Eq. (1).

Some ODE software systems can automatically locate and classify critical points. For instance, Fig. 9.2.17 shows a screen produced by the Matlab program pplane that is cited in the Section 9.1 application. It shows that the fourth-quadrant critical point in Fig. 9.2.16 has approximate coordinates (1.5708, 2.1416), and that the coefficient matrix of the associated linear system has the positive eigenvalue λ12.8949 and the negative eigenvalue λ22.3241. It therefore follows from Theorem 2 that this critical point is, indeed, a saddle point of the almost linear system in (1).

FIGURE 9.2.17.

The fourth-quadrant saddle point revealed.

With a general computer algebra system such as Maple or Mathematica, you may have to do a bit of work yourself—or tell the computer precisely what to do—to find and classify a critical point. For instance, the Maple command


fsolve({−y∗cos(x+y−1)=0,x∗cos(x−y+1)=0},
    {x,y},{x=1..2,y=−3..−2});

or the Mathematica command

FindRoot[{−y∗Cos[x+y−1] == 0, x∗Cos[x−y+1] == 0},
     {x,1,2}, {y,−3,−2}]

will find the critical-point coordinates a=1.5708, b=2.1416 indicated earlier. Similarly, Fig. 9.2.18 shows a TI handheld calculation of this critical point. Then the substitution x=u+a, y=v+b yields the translated system

(2)dudt=(2.1416v)cos(1.5708uv)=f(u,v),dvdt=(1.5708+u)cos(4.7124+uv)=g(u,v).

FIGURE 9.2.18.

TI-Nspire CX CAS calculation of the fourth-quadrant critical point of the almost linear system (1).

If we substitute u=v=0 in the Jacobian matrix (f,g)/(u,v), we get the coefficient matrix

A=[2.14162.14161.57081.5708]

of the linear system corresponding to (2). Then the Maple command


evalf(Eigenvals(A))

or the Mathematica command


Eigenvalues[A]

or the Wolfram|Alpha query


((2.1416, 2.1416), (1.5708, 1.5708))

yields the eigenvalues λ12.8949 and λ22.3241, thereby verifying that the critical point (1.5708,2.1416) of (1) is, indeed, a saddle point.

Use a computer algebra system to find and classify the other critical points of (1) indicated in Fig. 9.2.16. Then investigate similarly an almost linear system of your own construction. One convenient way to construct such a system is to begin with a linear system and insert sine or cosine factors resembling the ones in (1).

9.3 Ecological Models: Predators and Competitors

Some of the most interesting and important applications of stability theory involve the interactions between two or more biological populations occupying the same environment. We consider first a predator–prey situation involving two species. One species—the predators—feeds on the other species—the prey—which in turn feeds on some third food item readily available in the environment. A standard example is a population of foxes and rabbits in a woodland; the foxes (predators) eat rabbits (the prey), while the rabbits eat certain vegetation in the woodland. Other examples are sharks (predators) and food fish (prey), bass (predators) and sunfish (prey), ladybugs (predators) and aphids (prey), and beetles (predators) and scale insects (prey).

The classical mathematical model of a predator–prey situation was developed in the 1920s by the Italian mathematician Vito Volterra (1860–1940) in order to analyze the cyclic variations observed in the shark and food-fish populations in the Adriatic Sea. To construct such a model, we denote the number of prey at time t by x(t), the number of predators by y(t), and make the following simplifying assumptions.

  1. In the absence of predators, the prey population would grow at a natural rate, with dx/dt=ax, a>0.

  2. In the absence of prey, the predator population would decline at a natural rate, with dy/dt=by, b>0.

  3. When both predators and prey are present, there occurs, in combination with these natural rates of growth and decline, a decline in the prey population and a growth in the predator population, each at a rate proportional to the frequency of encounters between individuals of the two species. We assume further that the frequency of such encounters is proportional to the product xy, reasoning that doubling either population alone should double the frequency of encounters, while doubling both populations ought to quadruple the frequency of encounters. Consequently, the consumption of prey by predators results in

    • an interaction rate of decline pxy in the prey population x, and

    • an interaction rate of growth qxy in the predator population y.

When we combine the natural and interaction rates ax and pxy for the prey population x, as well as the natural and interaction rates by and qxy for the predator population y, we get the predator–prey system

(1)dxdt=axpxy=x(apy),dydt=by+qxy=y(b+qx),

with the constants a, b, p, and q all positive. [Note: You may see the predator and prey equations written in either order in (1). It is important to recognize that the predator equation has negative linear term and positive interaction term, whereas the prey equation has positive linear term and negative interaction term.]

Example 1

The critical points A critical point of the general predator–prey system in (1) is a solution (x, y) of the equations

(2)x(apy)=0,y(b+qx)=0.

The first of these two equations implies that either x=0 or y=a/p0, and the second implies that either y=0 or x=b/q0. It follows readily that this predator–prey system has the two (isolated) critical points (0, 0) and (b/q, a/p).

The Critical Point (0, 0): The Jacobian matrix of the system in (1) is

(3)J(x,y)=[apypxqyb+qx],soJ(0,0)=[a00b].

The matrix J(0, 0) has characteristic equation (aλ)(bλ)=0 and the eigenvalues λ1=a>0, λ2=b<0 with different signs. Hence it follows from Theorems 1 and 2 in Section 9.2 that the critical point (0, 0) is an unstable saddle point, both of the predator–prey system and of its linearization at (0, 0). The corresponding equilibrium solution x(t)0, y(t)0 merely describes simultaneous extinction of the prey (x) and predator (y) populations.

The Critical Point (b/q, a/p): The Jacobian matrix

(4)J(b/q,a/p)=[0pbqaqp0]

has characteristic equation λ2+ab=0 and the pure imaginary eigenvalues λ1, λ2=±iab. It follows from Theorem 1 in Section 9.2 that the linearization of the predator–prey system at (b/q, a/p) has a stable center at the origin. Thus we have the indeterminate case of Theorem 2 in Section 9.2, in which case the critical point can (aside from a stable center) also be either a stable spiral sink or an unstable spiral source of the predator–prey system itself. Hence further investigation is required to determine the actual character of the critical point (b/q, a/p). The corresponding equilibrium solution x(t)b/q, y(t)a/p describes the only nonzero constant prey (x) and predator (y) populations that coexist permanently.

The Phase Plane Portrait: In Problem 1 we ask you to analyze numerically a typical predator–prey system and verify that the linearizations at its two critical points agree qualitatively with the phase plane portrait shown in Fig. 9.3.1—where the nontrivial critical point appears visually to be a stable center. Of course, only the first quadrant of this portrait corresponds to physically meaningful solutions describing nonnegative populations of prey and predators.

In Problem 2 we ask you to derive an exact implicit solution of the predator–prey system of Fig. 9.3.1—a solution that can be used to show that its phase plane trajectories in the first quadrant are, indeed, simple closed curves that encircle the critical point (75, 50) as indicated in the figure. It then follows from Problem 30 in Section 9.1 that the explicit solution functions x(t) and y(t) are both periodic functions of t—thus explaining the periodic fluctuations that are observed empirically in predator–prey populations.

FIGURE 9.3.1.

Phase plane portrait for the predator–prey system x=200x4xy,y=150y+2xy with critical points (0, 0) and (75, 50).

FIGURE 9.3.2.

The predator–prey phase portrait of Example 2.

Example 2

Oscillating populations Figure 9.3.2 shows a computer-generated direction field and phase portrait for the predator–prey system

(5)dxdt=(0.2)x(0.005)xy=(0.005)x(40y),dydt=(0.5)y+(0.01)xy=(0.01)y(50+x),

where x(t) denotes the number of rabbits and y(t) the number of foxes after t months. Evidently the critical point (50, 40) is a stable center representing equilibrium populations of 50 rabbits and 40 foxes. Any other initial point lies on a closed trajectory enclosing this equilibrium point. The direction field indicates that the point (x(t), y(t)) traverses its trajectory in a counterclockwise direction, with the rabbit and fox populations oscillating periodically between their separate maximum and minimum values. A drawback is that the phase plane plot provides no indication as to the speed with which each trajectory is traversed.

This lost “sense of time” is recaptured by graphing the two individual population functions as functions of time t. In Fig. 9.3.3 we have graphed approximate solution functions x(t) and y(t) calculated using the Runge-Kutta method of Section 7.7 with initial values x(0)=70 and y(0)=40. We see that the rabbit population oscillates between the extreme values xmax72 and xmin33, while the fox population oscillates (out of phase) between the extreme values ymax70 and ymin20. A careful measurement indicates that the period P of oscillation of each population is slightly over 20 months. One could “zoom in” on the maximum/minimum points on each graph in order to refine these estimates of the period and the maximum and minimum rabbit and fox populations.

FIGURE 9.3.3.

Periodic oscillations of the predator and prey populations in Example 2.

Any positive initial conditions x0=x(0) and y0=y(0) yield a similar picture, with the rabbit and fox populations both surviving in coexistence with each other.

Competing Species

Now we consider two species (of animals, plants, or bacteria, for instance) with populations x(t) and y(t) at time t that compete with each other for the food available in their common environment. This is in marked contrast to the case in which one species preys on the other. To construct a mathematical model that is as realistic as possible, let us assume that in the absence of either species, the other would have a bounded (logistic) population like those considered in Section 2.1. In the absence of any interaction or competition between the two species, their populations x(t) and y(t) would then satisfy the differential equations

(6)dxdt=a1xb1x2,dydt=a2yb2y2,

each of the form of Eq. (2) of Section 2.1. But in addition, we assume that competition has the effect of a rate of decline in each population that is proportional to their product xy. We insert such terms with negative proportionality constants c1 and c2 in the equations in (6) to obtain the competition system

(7)dxdt=a1xb1x2c1xy=x(a1b1xc1y),dydt=a2yb2y2c2xy=y(a2b2yc2x),

where the coefficients a1, a2, b1, b2, c1, and c2 are all positive.

The almost linear system in (7) has four critical points. Upon setting the right-hand sides of the two equations equal to zero, we see that if x=0, then either y=0 or y=a2/b2, whereas if y=0, then either x=0 or x=a1/b1. This gives the three critical points (0,0), (0,a2/b2), and (a1/b1,0). The fourth critical point is obtained from the simultaneous solution of the equations

(8)b1x+c1y=a1,c2x+b2y=a2.

We assume that, as in most interesting applications, these equations have a single solution and that the corresponding critical point lies in the first quadrant of the xy-plane. This point (xE,yE) is then the fourth critical point of the system in (7), and it represents the possibility of coexistence of the two species, with constant nonzero equilibrium populations x(t)xE and y(t)yE.

We are interested in the stability of the critical point (xE,yE). This turns out to depend on whether

(9)c1c2<b1b2orc1c2>b1b2.

Each inequality in (9) has a natural interpretation. Examining the equations in (6), we see that the coefficients b1 and b2 represent the inhibiting effect of each population on its own growth (possibly due to limitations of food or space). On the other hand, c1 and c2 represent the effect of competition between the two populations. Thus b1b2 is a measure of inhibition, while c1c2 is a measure of competition. A general analysis of the system in (7) shows the following:

  1. If c1c2<b1b2, so that competition is small in comparison with inhibition, then (xE,yE) is an asymptotically stable critical point that is approached by each solution as t+. Thus the two species can and do coexist in this case.

  2. If c1c2>b1b2, so that competition is large in comparison with inhibition, then (xE,yE) is an unstable critical point, and either x(t) or y(t) approaches zero as t+. Thus the two species cannot coexist in this case; one survives and the other becomes extinct.

Rather than carrying out this general analysis, we present two examples that illustrate these two possibilities.

Example 3

Survival of a single species Suppose that the populations x(t) and y(t) satisfy the equations

(10)dxdt=14x12x2xy,dydt=16y12y2xy,

in which a1=14, a2=16, b1=b2=12, and c1=c2=1. Then c1c2=1>14=b1b2, so we should expect survival of a single species as predicted in Case 2 previously. We find readily that the four critical points are (0, 0), (0, 32), (28, 0), and (12, 8). We shall investigate them individually.

The Critical Point (0, 0): The Jacobian matrix of the system in (10) is

(11)J(x,y)=[14xyxy16yx],soJ(0,0)=[140016].

The matrix J(0, 0) has characteristic equation (14λ)(16λ)=0 and has the eigenvalues

λ1=14with eigenvectorv1=[10]T

and

λ2=16with eigenvectorv2=[01]T.

Both eigenvalues are positive, so it follows that (0, 0) is a nodal source for the system’s linearization x=14x, y=16y at (0, 0), and hence—by Theorem 2 in Section 9.2—is also an unstable nodal source for the original system in (10). Figure 9.3.4 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.4.

Phase plane portrait for the linear system x=14x, y=16y corresponding to the critical point (0, 0).

The Critical Point (0, 32): Substitution of x=0, y=32 in the Jacobian matrix J(x, y) shown in (11) yields the Jacobian matrix

(12)J(0,32)=[1803216]

of the nonlinear system (10) at the point (0, 32). Comparing Eqs. (7) and (8) in Section 9.2, we see that this Jacobian matrix corresponds to the linearization

(13)dudt=18u,dvdt=32u16v

of (10) at (0, 32). The matrix J(0, 32) has characteristic equation (18λ)(16λ)=0 and has the eigenvalues λ1=18 with eigenvector v1=[116]T and λ2=16 with eigenvector v2=[01]T. Because both eigenvalues are negative, it follows that (0, 0) is a nodal sink for the linearized system, and hence—by Theorem 2 in Section 9.2—that (0, 32) is also a stable nodal sink for the original system in (10). Figure 9.3.5 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.5.

Phase plane portrait for the linear system in Eq. (13) corresponding to the critical point (0, 32).

The Critical Point (28, 0): The Jacobian matrix

(14)J(28,0)=[1428012]

corresponds to the linearization

(15)dudt=14u28v,dvdt=12v

of (10) at (28, 0). The matrix J(28, 0) has characteristic equation (14λ)(12λ)=0 and has the eigenvalues λ1=14 with eigenvector v1=[10]T and λ2=12 with eigenvector v2=[141]T. Because both eigenvalues are negative, it follows that (0, 0) is a nodal sink for the linearized system, and hence—by Theorem 2 in Section 9.2—that (28, 0) is also a stable nodal sink for the original nonlinear system in (10). Figure 9.3.6 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.6.

Phase plane portrait for the linear system in Eq. (15) corresponding to the critical point (28, 0).

The Critical Point (12, 8): The Jacobian matrix

(16)J(12,8)=[61284]

corresponds to the linearization

(17)dudt=6u12v,dvdt=8u4v

of (10) at (12, 8). The matrix J(12, 8) has characteristic equation

(6λ)(4λ)(8)(12)=λ2+10λ72=0

and has the eigenvalues

λ1=597<0with eigenvectorv1=[18(1+97)1]T

and

λ2=5+97>0with eigenvectorv2=[18(1+97)1]T.

Because the two eigenvalues have opposite signs, it follows that (0, 0) is a saddle point for the linearized system and hence—by Theorem 2 in Section 9.2—that (12, 8) is also an unstable saddle point for the original system in (10). Figure 9.3.7 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.7.

Phase plane portrait for the linear system in Eq. (17) corresponding to the critical point (12, 8).

Now that our local analysis of each of the four critical points is complete, it remains to assemble the information found into a coherent global picture. If we accept the facts that

  • Near each critical point, the trajectories for the original system in (10) resemble qualitatively the linearized trajectories shown in Figs. 9.3.49.3.7, and

  • As t+ each trajectory either approaches a critical point or diverges toward infinity,

then it would appear that the phase plane portrait for the original system must resemble the rough sketch shown in Fig. 9.3.8. This sketch shows a few typical freehand trajectories connecting a nodal source at (0, 0), nodal sinks at (0, 32) and (28, 0), and a saddle point at (12, 8), with indicated directions of flow along these trajectories consistent with the known character of these critical points. Figure 9.3.9 shows a more precise computer-generated phase portrait and direction field for the nonlinear system in (10).

FIGURE 9.3.8.

Rough sketch consistent with the analysis in Example 3.

FIGURE 9.3.9.

Phase plane portrait for the system in Example 3.

The two trajectories that approach the saddle point (12, 8), together with that saddle point, form a separatrix that separates Regions I and II in Fig. 9.3.8. It plays a crucial role in determining the long-term behavior of the two populations. If the initial point (x0,y0) lies precisely on the separatrix, then (x(t), y(t)) approaches (12, 8) as t+. Of course, random events make it extremely unlikely that (x(t), y(t)) will remain on the separatrix. If not, peaceful coexistence of the two species is impossible. If (x0,y0) lies in Region I above the separatrix, then (x(t), y(t)) approaches (0, 32) as t+, so the population x(t) decreases to zero. Alternatively, if (x0,y0) lies in Region II below the separatrix, then (x(t), y(t)) approaches (28, 0) as t+, so the population y(t) dies out. In short, whichever population has the initial competitive advantage survives, while the other faces extinction.

Example 4

Peaceful coexistence of two species Suppose that the populations x(t) and y(t) satisfy the competition system

(18)dxdt=14x2x2xy,dydt=16y2y2xy,

for which a1=14, a2=16, b1=b2=2, and c1=c2=1. Then c1c2=1<4=b1b2, so now the effect of inhibition is greater than that of competition. We find readily that the four critical points are (0, 0), (0, 8), (7, 0), and (4, 6). We proceed as in Example 3.

The Critical Point (0, 0): When we drop the quadratic terms in (18), we get the same linearization x=14x, y=16y at (0, 0) as in Example 3. Thus its coefficient matrix has the two positive eigenvalues λ1=14 and λ2=16, and its phase portrait is the same as that shown in Fig. 9.3.4. Therefore, (0, 0) is an unstable nodal source for the original system in (18).

The Critical Point (0, 8): The Jacobian matrix of the system in (18) is

(19)J(x,y)=[144xyxy164yx],soJ(0,8)=[60816].

The matrix J(0, 8) corresponds to the linearization

(20)dudt=6u,dvdt=8u16v

of (18) at (0, 8). It has characteristic equation (6λ)(16λ)=0 and has the positive eigenvalue λ1=6 with eigenvector v1=[114]T and the negative eigenvalue λ2=16 with eigenvector v2=[01]T. It follows that (0, 0) is a saddle point for the linearized system, and hence that (0, 8) is an unstable saddle point for the original system in (18). Figure 9.3.10 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.10.

Phase plane portrait for the linear system in Eq. (20) corresponding to the critical point (0, 8).

The Critical Point (7, 0): The Jacobian matrix

(21)J(7,0)=[14709]

corresponds to the linearization

(22)dudt=14u7v,dvdt=9v

of (18) at (7, 0). The matrix J(7, 0) has characteristic equation (14λ)(9λ)=0 and has the negative eigenvalue λ1=14 with eigenvector v1=[10]T and the positive eigenvalue λ2=9 with eigenvector v2=[723]T. It follows that (0, 0) is a saddle point for the linearized system, and hence that (7, 0) is an unstable saddle point for the original system in (18). Figure 9.3.11 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.11.

Phase plane portrait for the linear system in Eq. (22) corresponding to the to the critical point (7, 0).

The Critical Point (4, 6): The Jacobian matrix

(23)J(4,6)=[84612]

corresponds to the linearization

(24)dudt=8u4v,dvdt=6u12v

of (18) at (4, 6). The matrix J(4, 6) has characteristic equation

(8λ)(12λ)(6)(4)=λ2+20λ+72=0

and has the two negative eigenvalues

λ1=2(57)with eigenvectorv1=[13(1+7)1]T

and

λ2=2(5+7)with eigenvectorv2=[13(17)1]T.

It follows that (0, 0) is a nodal sink for the linearized system, and hence that (4, 6) is a stable nodal sink for the original system in (18). Figure 9.3.12 shows a phase portrait for the linearized system near (0, 0).

FIGURE 9.3.12.

Phase plane portrait for the linear system in Eq. (24) corresponding to the critical point (4, 6).

Figure 9.3.13 assembles all this local information into a global phase plane portrait for the original system in (18). The notable feature of this system is that—for any positive initial population values x0 and y0—the point (x(t), y(t)) approaches the single critical point (4, 6) as t+. It follows that the two species both survive in stable (peaceful) existence.

Interactions of Logistic Populations

If the coefficients a1, a2, b1, b2 are positive but c1=c2=0, then the equations

(25)dxdt=a1xb1x2c1xy,dydt=a2yb2y2c2xy

describe two separate logistic populations x(t) and y(t) that have no effect on each other. Examples 3 and 4 illustrate cases in which the xy-coefficients c1 and c2 are both positive. The interaction between the two populations is then described as competition, because the effect of the xy-terms in (25) is to decrease the rates of growth of both populations—that is, each population is “hurt” by their mutual interaction.

Suppose, however, that the interaction coefficients c1 and c2 in (25) are both negative. Then the effect of the xy-terms is to increase the rates of growth of both populations—that is, each population is “helped” by their mutual interaction. This type of interaction is aptly described as cooperation between the two logistic populations.

FIGURE 9.3.13.

Direction field and phase portrait for the competition system x=14x2x2xy,y=16y2y2xy of Example 4.

Finally, the interaction between the two populations is one of predation if the interaction coefficients have different signs. For instance, if c1>0 but c2<0, then the x-population is hurt but the y-population is helped by their interaction. We may therefore describe x(t) as a prey population and y(t) as a predator population.

If either b1 or b2 is zero in (25), then the corresponding population would (in the absence of the other) exhibit exponential growth rather than logistic growth. For instance, suppose that a1>0, a2<0, b1=b2=0, and c1>0, c2<0. Then x(t) is a naturally growing prey population while y(t) is a naturally declining predator population. This is the original predator–prey model with which we began this section.

Problems 26 through 34 illustrate a variety of the possibilities indicated here. The problems and examples in this section illustrate the power of elementary critical-point analysis. But remember that ecological systems in nature are seldom so simple as in these examples. Frequently they involve more than two species, and the growth rates of these populations and the interactions among them often are more complicated than those discussed in this section. Consequently, the mathematical modeling of ecological systems remains an active area of current research.

9.3 Problems

Predator–Prey System

Problems 1 and 2 deal with the predator–prey system

(1)dxdt=200x4xy,dydt=150y+2xy

that corresponds to Fig. 9.3.1.

  1. Starting with the Jacobian matrix of the system in (1), derive its linearizations at the two critical points (0, 0) and (75, 50). Use a graphing calculator or computer system to construct phase plane portraits for these two linearizations that are consistent with the “big picture” shown in Fig. 9.3.1.

  2. Separate the variables in the quotient

    dydx=150y+2xy200x4xy

    of the two equations in (1), and thereby derive the exact implicit solution

    200ln y+150ln x2x4y=C

    of the system. Use the contour plot facility of a graphing calculator or computer system to plot the contour curves of this equation through the points (75, 100), (75, 150), (75, 200), (75, 250), and (75, 300) in the xy-plane. Are your results consistent with Fig. 9.3.1?

  3. Insect population Let x(t) be a harmful insect population (aphids?) that under natural conditions is held somewhat in check by a benign predator insect population y(t) (ladybugs?). Assume that x(t) and y(t) satisfy the predator–prey equations in (1), so that the stable equilibrium populations are xE=b/q and yE=a/p. Now suppose that an insecticide is employed that kills (per unit time) the same fraction f<a of each species of insect. Show that the harmful population xE is increased, while the benign population yE is decreased, so the use of the insecticide is counterproductive. This is an instance in which mathematical analysis reveals undesirable consequences of a well-intentioned interference with nature.

Competition System

Problems 4 through 7 deal with the competition system

(2)dxdt=60x4x23xy,dydt=42y2y23xy,

in which c1c2=9>8=b1b2, so the effect of competition should exceed that of inhibition. Problems 4 through 7 imply that the four critical points (0, 0), (0, 21), (15, 0), and (6, 12) of the system in (2) resemble those shown in Fig. 9.3.9—a nodal source at the origin, a nodal sink on each coordinate axis, and a saddle point interior to the first quadrant. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Finally, construct a first-quadrant phase plane portrait for the nonlinear system in (2). Do your local and global portraits look consistent?

  1. Show that the coefficient matrix of the linearization x=60x, y=42y of (2) at (0, 0) has positive eigenvalues λ1=60 and λ2=42. Hence (0, 0) is a nodal source for (2).

  2. Show that the linearization of (2) at (0, 21) is u=3u, v=63u42v. Then show that the coefficient matrix of this linear system has negative eigenvalues λ1=3 and λ2=42. Hence (0, 21) is a nodal sink for the system in (2).

  3. Show that the linearization of (2) at (15, 0) is u=60u45v,v=3v. Then show that the coefficient matrix of this linear system has negative eigenvalues λ1=60 and λ2=3. Hence (15, 0) is a nodal sink for the system in (2).

  4. Show that the linearization of (2) at (6, 12) is u=24u18v,v=36u24v. Then show that the coefficient matrix of this linear system has eigenvalues λ1=24182<0 and λ2=24+182>0. Hence (6, 12) is a saddle point for the system in (2).

Competition System

Problems 8 through 10 deal with the competition system

(3)dxdt=60x3x24xy,dydt=42y3y22xy,

in which c1c2=8<9=b1b2, so the effect of inhibition should exceed that of competition. The linearization of the system in (3) at (0, 0) is the same as that of (2). This observation and Problems 8 through 10 imply that the four critical points (0, 0), (0, 14), (20, 0), and (12, 6) of (3) resemble those shown in Fig. 9.3.13—a nodal source at the origin, a saddle point on each coordinate axis, and a nodal sink interior to the first quadrant. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Finally, construct a first-quadrant phase plane portrait for the nonlinear system in (3). Do your local and global portraits look consistent?

  1. Show that the linearization of (3) at (0, 14) is u=4u, v=28u42v. Then show that the coefficient matrix of this linear system has the positive eigenvalue λ1=4 and the negative eigenvalue λ2=42. Hence (0, 14) is a saddle point for the system in (3).

  2. Show that the linearization of (3) at (20, 0) is u=60u80v, v=2v. Then show that the coefficient matrix of this linear system has the negative eigenvalue λ1=60 and the positive eigenvalue λ2=2. Hence (20, 0) is a saddle point for the system in (3).

  3. Show that the linearization of (3) at (12, 6) is u=36u48v, v=12u18v. Then show that the coefficient matrix of this linear system has eigenvalues λ1=27+373 and λ2=27373, both of which are negative. Hence (12, 6) is a nodal sink for the system in (3).

Logistic Prey Population

Problems 11 through 13 deal with the predator–prey system

(4)dxdt=5xx2xy,dydt=2y+xy,

in which the prey population x(t) is logistic but the predator population y(t) would (in the absence of any prey) decline naturally. Problems 11 through 13 imply that the three critical points (0, 0), (5, 0), and (2, 3) of the system in (4) are as shown in Fig. 9.3.14—with saddle points at the origin and on the positive x-axis, and with a spiral sink interior to the first quadrant. In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.14?

FIGURE 9.3.14.

Direction field and phase portrait for the predator–prey system of Problems 11 through 13.

  1. Show that the coefficient matrix of the linearization x=5x, y=2y of (4) at (0, 0) has the positive eigenvalue λ1=5 and the negative eigenvalue λ2=2. Hence (0, 0) is a saddle point of the system in (4).

  2. Show that the linearization of (4) at (5, 0) is u=5u5v, v=3v. Then show that the coefficient matrix of this linear system has the negative eigenvalue λ1=5 and the positive eigenvalue λ2=3. Hence (5, 0) is a saddle point for the system in (4).

  3. Show that the linearization of (4) at (2, 3) is u=2u2v, v=3u. Then show that the coefficient matrix of this linear system has the complex conjugate eigenvalues λ1,  λ2=1±i5 with negative real part. Hence (2, 3) is a spiral sink for the system in (4).

Doomsday vs. Extinction

Problems 14 through 17 deal with the predator–prey system

(5)dxdt=x22xxy,dydt=y24y+xy.

Here each population—the prey population x(t) and the predator population y(t)—is an unsophisticated population (like the alligators of Section 2.1) for which the only alternatives (in the absence of the other population) are doomsday and extinction. Problems 14 through 17 imply that the four critical points (0, 0), (0, 4), (2, 0), and (3, 1) of the system in (5) are as shown in Fig. 9.3.15—a nodal sink at the origin, a saddle point on each coordinate axis, and a spiral source interior to the first quadrant. This is a two-dimensional version of “doomsday versus extinction.” If the initial point (x0,y0) lies in Region I, then both populations increase without bound (until doomsday), whereas if it lies in Region II, then both populations decrease to zero (and thus both become extinct). In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.15?

FIGURE 9.3.15.

Direction field and phase portrait for the predator–prey system of Problems 14 through 17.

  1. Show that the coefficient matrix of the linearization x=2x, y=4y of the system in (5) at (0, 0) has the negative eigenvalues λ1=2 and λ2=4. Hence (0, 0) is a nodal sink for (5).

  2. Show that the linearization of (5) at (0, 4) is u=6u, v=4u+4v. Then show that the coefficient matrix of this linear system has the negative eigenvalue λ1=6 and the positive eigenvalue λ2=4. Hence (0, 4) is a saddle point for the system in (5).

  3. Show that the linearization of (5) at (2, 0) is u=2u2v, v=2v. Then show that the coefficient matrix of this linear system has the positive eigenvalue λ1=2 and the negative eigenvalue λ2=2. Hence (2, 0) is a saddle point for the system in (5).

  4. Show that the linearization of (5) at (3, 1) is u=3u3v, v=u+v. Then show that the coefficient matrix of this linear system has complex conjugate eigenvalues λ1, λ2=2±i2 with positive real part. Hence (3, 1) is a spiral source for (5).

Problems 18 through 25 deal with the predator–prey system

(6)dxdt=2xxy+ϵx(5x),dydt=5y+xy,

for which a bifurcation occurs at the value ϵ=0 of the parameter ϵ. Problems 18 and 19 deal with the case ϵ=0, in which case the system in (6) takes the form

(7)dxdt=2xxy,dydt=5x+xy,

and these problems suggest that the two critical points (0, 0) and (5, 2) of the system in (7) are as shown in Fig. 9.3.16—a saddle point at the origin and a center at (5, 2). In each problem use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.16?

FIGURE 9.3.16.

The case ϵ=0 (Problems 18 and 19).

  1. Show that the coefficient matrix of the linearization x=2x, y=5y of (7) at (0, 0) has the positive eigenvalue λ1=2 and the negative eigenvalue λ2=5. Hence (0, 0) is a saddle point for the system in (7).

  2. Show that the linearization of the system in (7) at (5, 2) is u=5v, v=2u. Then show that the coefficient matrix of this linear system has conjugate imaginary eigenvalues λ1,  λ2=±i10. Hence (0, 0) is a stable center for the linear system. Although this is the indeterminate case of Theorem 2 in Section 9.2, Fig. 9.3.16 suggests that (5, 2) also is a stable center for (7).

Problems 20 through 22 deal with the case ϵ=1, for which the system in (6) becomes

(8)dxdt=3x+x2xy,dydt=5y+xy,

and imply that the three critical points (0, 0), (3, 0), and (5, 2) of (8) are as shown in Fig. 9.3.17—with a nodal sink at the origin, a saddle point on the positive x-axis, and a spiral source at (5, 2). In each problem use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.17?

FIGURE 9.3.17.

The case ϵ=1 (Problems 20 through 22).

  1. Show that the coefficient matrix of the linearization x=3x, y=5y of the system in (8) at (0, 0) has the negative eigenvalues λ1=3 and λ2=5. Hence (0, 0) is a nodal sink for (8).

  2. Show that the linearization of the system in (8) at (3, 0) is u=3u3v, v=2v. Then show that the coefficient matrix of this linear system has the positive eigenvalue λ1=3 and the negative eigenvalue λ2=2. Hence (3, 0) is a saddle point for (8).

  3. Show that the linearization of (8) at (5, 2) is u=5u5v, v=2u. Then show that the coefficient matrix of this linear system has complex conjugate eigenvalues λ1, λ2=12(5±i15) with positive real part. Hence (5, 2) is a spiral source for the system in (8).

Problems 23 through 25 deal with the case ϵ=1, so that the system in (6) takes the form

(9)dxdt=7xx2xy,dydt=5y+xy,

and these problems imply that the three critical points (0, 0), (7, 0), and (5, 2) of the system in (9) are as shown in Fig. 9.3.18—with saddle points at the origin and on the positive x-axis and with a spiral sink at (5, 2). In each problem use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.18?

FIGURE 9.3.18.

The case ϵ=+1 (Problems 23 through 25).

  1. Show that the coefficient matrix of the linearization x=7x, y=5y of (9) at (0, 0) has the positive eigenvalue λ1=7 and the negative eigenvalue λ2=5. Hence (0, 0) is a saddle point for the system in (9).

  2. Show that the linearization of (9) at (7, 0) is u=7u7v, v=2v. Then show that the coefficient matrix of this linear system has the negative eigenvalue λ1=7 and the positive eigenvalue λ2=2. Hence (7, 0) is a saddle point for the system in (9).

  3. Show that the linearization of (9) at (5, 2) is u=5u5v, v=2u. Then show that the coefficient matrix of this linear system has the complex conjugate eigenvalues λ1, λ2=12(5±i15) with negative real part. Hence (5, 2) is a spiral sink for the system in (9).

For each two-population system in Problems 26 through 34, first describe the type of x- and y-populations involved (exponential or logistic) and the nature of their interaction—competition, cooperation, or predation. Then find and characterize the system’s critical points (as to type and stability). Determine what nonzero x- and y-populations can coexist. Finally, construct a phase plane portrait that enables you to describe the long-term behavior of the two populations in terms of their initial populations x(0) and y(0).

  1. dxdt=2xxy,dydt=3yxy

  2. dxdt=2xy4x,dydt=xy3y

  3. dxdt=2xy16x,dydt=4yxy

  4. dxdt=3xx212xy,dydt=4y2xy

  5. dxdt=3xx2+12xy,dydt=15xyy

  6. dxdt=3xx214xy,dydt=xy2y

  7. dxdt=30x3x2+xy,dydt=60y3y2+4xy

  8. dxdt=30x2x2xy,dydt=80y4y2+2xy

  9. dxdt=30x2x2xy,dydt=20y4y2+2xy

9.3 Application Your Own Wildlife Conservation Preserve

You own a large wildlife conservation preserve that you originally stocked with F0 foxes and R0 rabbits on January 1, 2007. The following differential equations model the numbers R(t) of rabbits and F(t) foxes t months later:

dRdt=(0.01)pR(0.0001)aRF,dFdt=(0.01)qF+(0.0001)bRF,

where p and q are the two largest digits (with p<q) and a and b are the two smallest nonzero digits (with a<b) in your student ID number.

The numbers of foxes and rabbits will oscillate periodically, out of phase with each other (like the functions x(t) and y(t) in Fig. 9.3.3). Choose your initial numbers F0 of foxes and R0 of rabbits—perhaps several hundred of each—so that the resulting solution curve in the RF-plane is a fairly eccentric closed curve. (The eccentricity may be increased if you begin with a relatively large number of rabbits and a small number of foxes, as any wildlife preserve owner would naturally do—because foxes prey on rabbits.)

Your task is to determine

  1. The period of oscillation of the rabbit and fox populations;

  2. The maximum and minimum numbers of rabbits, and the calendar dates on which they first occur;

  3. The maximum and minimum numbers of foxes, and the calendar dates on which they first occur.

With computer software that can plot both RF-trajectories and tR- and tF-solution curves like those in Figs. 9.3.2 and 9.3.3, you can “zoom in” graphically on the points whose coordinates provide the requested information.

9.4 Nonlinear Mechanical Systems

Now we apply the qualitative methods of Sections 9.1 and 9.2 to the analysis of simple mechanical systems like the mass-on-a-spring system shown in Fig. 9.4.1. Let m denote the mass in a suitable system of units and let x(t) denote the displacement of the mass at time t from its equilibrium position (in which the spring is unstretched). Previously we have always assumed that the force F(x) exerted by the spring on the mass is a linear function of x: F(x)=kx (Hooke’s law). In reality, however, every spring in nature actually is nonlinear (even if only slightly so). Moreover, springs in some automobile suspension systems deliberately are designed to be nonlinear. Here, then, we are interested specifically in the effects of nonlinearity.

FIGURE 9.4.1.

The mass on a spring.

So now we allow the force function F(x) to be nonlinear. Because F(0)=0 at the equilibrium position x=0, we may assume that F has a power series expansion of the form

(1)F(x)=kx+αx2+βx3+.

We take k>0 so that the reaction of the spring is directed opposite to the displacement when x is sufficiently small. If we assume also that the reaction of the spring is symmetric with respect to positive and negative displacements by the same distance, then F(x)=F(x), so F is an odd function. In this case it follows that the coefficient of xn in Eq. (1) is zero if n is even, so the first nonlinear term is the one involving x3.

For a simple mathematical model of a nonlinear spring we therefore take

(2)F(x)=kx+βx3,

ignoring all terms in Eq. (1) of degree greater than 3. The equation of motion of the mass m is then

(3)mx=kx+βx3.

The Position–Velocity Phase Plane

If we introduce the velocity

(4)y(t)=x(t)

of the mass with position x(t), then we get from Eq. (3) the equivalent first-order system

(5)dxdt=y,mdydt=kx+βx3.

A phase plane trajectory of this system is a position-velocity plot that illustrates the motion of the mass on the spring. We can solve explicitly for the trajectories of this system by writing

dydx=dy/dtdx/dt=kx+βx3my,

whence

my dy+(kxβx3)dx=0.

Integration then yields

(6)12my2+12kx214βx4=E

for the equation of a typical trajectory. We write E for the arbitrary constant of integration because KE=12my2 is the kinetic energy of the mass with velocity y, and it is natural to define

(7)PE=12kx214βx4

as the potential energy of the spring. Then Eq. (6) takes the form KE+PE=E, so the constant E turns out to be the total energy of the mass-spring system. Equation (6) then expresses conservation of energy for the undamped motion of a mass on a spring.

The behavior of the mass depends on the sign of the nonlinear term in Eq. (2). The spring is called

We consider the two cases separately.

Hard Spring Oscillations: If β<0, then the second equation in (5) takes the form my=x(|β|x2+k), so it follows that the only critical point of the system is the origin (0, 0). Each trajectory

(8)12my2+12kx214|β|x4=E>0

is an oval closed curve like those shown in Fig. 9.4.2, and thus (0, 0) is a stable center. As the point (x(t), y(t)) traverses a trajectory in the clockwise direction, the position x(t) and velocity y(t) of the mass oscillate alternately, as illustrated in Fig. 9.4.3. The mass is moving to the right (with x increasing) when y>0, to the left when y<0. Thus the behavior of a mass on a nonlinear hard spring resembles qualitatively that of a mass on a linear spring with β=0 (as in Example 3 of Section 9.1). But one difference between the linear and nonlinear situations is that, whereas the period T=2πm/k of oscillation of a mass on a linear spring is independent of the initial conditions, the period of a mass on a nonlinear spring depends on its initial position x(0) and initial velocity y(0) (Problems 21 through 26).

FIGURE 9.4.2.

Position-velocity phase plane portrait for the hard mass-and-spring system with m=k=2 and β=4<0.

FIGURE 9.4.3.

Position and velocity solution curves for the hard mass-and-spring system with m=k=2 and β=4<0.

Remark

The hard spring equation mx=kx|β|x3 has equivalent first-order system

x=y,y=kmx|β|mx3

with Jacobian matrix

J(x,y)=[01km3|β|mx20],soJ(0,0)=[01ω20]

(writing k/m=ω2 as usual). The latter matrix has characteristic equation λ2+ω2=0 and pure imaginary eigenvalues λ1, λ2=±ωi. Thus the linearized system x=y, y=ω2x has a stable center at the critical point (0, 0)—as we observed in Example 4 of Section 9.1. However, the nonlinear cubic term in the differential equation has (in effect) replaced the elliptical trajectories (as in Fig. 9.1.8) of the linear system with the “flatter” quartic ovals we see in Fig. 9.4.2.

Soft Spring Oscillations: If β>0, then the second equation in (5) takes the form my=x(βx2k), so it follows that the system has the two critical points (±k/β,0) in addition to the critical point (0, 0). These three critical points yield the only solutions for which the mass can remain at rest. The following example illustrates the greater range of possible behaviors of a mass on a soft spring.

Example 1

Undamped soft spring If m=1, k=4, and β=1, then the equation of motion of the mass is

(9)d2xdt2+4xx3=0,

and Eq. (6) gives the trajectories in the form

(10)12y2+2x214x4=E.

After solving for

(10′)y=±2E4x2+12x4,

we could select a fixed value of the constant energy E and plot manually a trajectory like one of those shown in the computer-generated position-velocity phase plane portrait in Fig. 9.4.4.

The different types of phase plane trajectories correspond to different values of the energy E. If we substitute x=±k/β and y=0 into (6), we get the energy value E=k2/(4β)=4 (because k=4 and β=1) that corresponds to the trajectories that intersect the x-axis at the nontrivial critical points (2,0) and (2, 0). These emphasized trajectories are called separatrices because they separate phase plane regions of different behavior.

The nature of the motion of the mass is determined by which type of trajectory its initial conditions determine. The simple closed trajectories encircling (0, 0) in the region bounded by the separatrices correspond to energies in the range 0<E<4. These closed trajectories represent periodic oscillations of the mass back and forth around the equilibrium point x=0.

The unbounded trajectories lying in the regions above and below the separatrices correspond to values of E greater than 4. These represent motions in which the mass approaches x=0 with sufficient energy that it continues on through the equilibrium point, never to return again (as indicated in Fig. 9.4.5).

FIGURE 9.4.4.

Position–velocity phase plane portrait for the soft mass-and-spring system with m=1,k=4, and β=1>0. The separatrices are emphasized.

FIGURE 9.4.5.

Position and velocity solution curves for the soft mass-and-spring system with m=1,k=4,β=1>0, and energy E=8—sufficiently great that the mass approaches the origin from the left and continues on indefinitely to the right.

The unbounded trajectories opening to the right and left correspond to values of E less than 4. These represent motions in which the mass initially is headed toward the equilibrium point x=0, but with insufficient energy to reach it. At some point the mass reverses direction and heads back whence it came.

In Fig. 9.4.4 it appears that the critical point (0, 0) is a stable center, whereas the critical points (±2,0) look like saddle points of the equivalent first-order system

(11)x=y,y=4x+x3

with Jacobian matrix

J(x,y)=[014+3x20].

To check these observations against the usual critical-point analysis, we note first that the Jacobian matrix

J(0,0)=[0140]

at the critical point (0, 0) has characteristic equation λ2+4=0 and pure imaginary eigenvalues λ1, λ2=±2i consistent with a stable center. Moreover, the Jacobian matrix

J(±2,0)=[0180]

corresponding to the other two critical points has characteristic equation λ28=0 and real eigenvalues λ1, λ2=±8 of opposite sign, consistent with the saddle-point behavior that we observe near (2,0) and (+2,0).

Remark

Figures 9.4.2 and 9.4.4 illustrate a significant qualitative difference between hard springs with β<0 and soft springs with β>0 in the nonlinear equation mx=kx+βx3. Whereas the phase plane trajectories for a hard spring are all bounded, a soft spring has unbounded phase plane trajectories (as well as bounded ones). However, we should realize that the unbounded soft-spring trajectories cease to represent physically realistic motions faithfully when they exceed the spring’s capability of expansion without breaking.

Damped Nonlinear Vibrations

Suppose now that the mass on a spring is connected also to a dashpot that provides a force of resistance proportional to the velocity y=dx/dt of the mass. If the spring is still assumed nonlinear as in Eq. (2), then the equation of motion of the mass is

(12)mx=cxkx+βx3,

where c>0 is the resistance constant. If β>0, then the equivalent first-order system

(13)dxdt=y,dydt=kxcy+βx3m=cmykmx(1βkx2)

has critical points (0, 0) and (±k/β,0) and Jacobian matrix

J(x,y)=[01km+3βmx2cm].

Now the critical point at the origin is the most interesting one. The Jacobian matrix

J(0,0)=[01kmcm]

has characteristic equation

(λ)(cmλ)+km=1m(mλ2+cλ+k)=0

and eigenvalues

λ1,λ2=c±c24km2m.

It follows from Theorem 2 in Section 9.2 in that the critical point (0, 0) of the system in (13) is

The following example illustrates the latter case. (In the borderline case with equal negative eigenvalues, the origin may be either a nodal or a spiral sink.)

Example 2

Damped soft spring Suppose that m=1, c=2, k=5, and β=54. Then the nonlinear system in (13) is

(14)dxdt=y,dydt=5x2y+54x3=2y5x(114x2).

It has critical points (0,0), (±2,0) and Jacobian matrix

J(x,y)=[015+154x22].

At (0, 0): The Jacobian matrix

J(0,0)=[0152]

has characteristic equation λ2+2λ+5=0 and has complex conjugate eigenvalues λ1, λ2=1±2i with negative real part. Hence (0, 0) is a spiral sink of the nonlinear system in (14), and the linearized position function of the mass is of the form

x(t)=et(Acos2t+Bsin2t)

that corresponds to an exponentially damped oscillation about the equilibrium position x=0.

At (±2,0): The Jacobian matrix

J(±2,0)=[01102]

has characteristic equation λ2+2λ10=0 and real eigenvalues λ1=111<0 and λ2=1+11>0 with different signs. It follows that (2,0) and (+2,0) are both saddle points of the system in (14).

The position–velocity phase plane portrait in Fig. 9.4.6 shows trajectories of (14) and the spiral sink at (0, 0), as well as the unstable saddle points at (2,0) and (2, 0). The emphasized separatrices divide the phase plane into regions of different behavior. The behavior of the mass depends on the region in which its initial point (x0,y0) is located. If this initial point lies in

  • Region I between the separatrices, then the trajectory spirals into the origin as t+, and hence the periodic oscillations of the undamped case (Fig. 9.4.4) are now replaced with damped oscillations around the stable equilibrium position x=0;

  • Region II, then the mass passes through x=0 moving from left to right (x increasing);

  • Region III, then the mass passes through x=0 moving from right to left (x decreasing);

  • Region IV, then the mass approaches (but does not reach) the unstable equilibrium position x=2 from the left, but stops and then returns to the left;

  • Region V, then the mass approaches (but does not reach) the unstable equilibrium position x=2 from the right, but stops and then returns to the right.

FIGURE 9.4.6.

Position–velocity phase plane portrait for the soft mass-and-spring system with m=1,k=5,β=54, and resistance constant c=2. The (black) separatrices are emphasized.

If the initial point (x0,y0) lies precisely on one of the separatrices, then the corresponding trajectory either approaches the stable spiral point or recedes to infinity from a saddle point as t+.

The Nonlinear Pendulum

In Section 5.4 we derived the equation

(15)d2θdt2+gLsinθ=0

for the undamped oscillations of the simple pendulum shown in Fig. 9.4.7. There we used the approximation sin θθ for θ near zero to replace Eq. (15) with the linear model

(16)d2θdt2+ω2θ=0,

where ω2=g/L. The general solution

(17)θ(t)=Acosωt+Bsinωt

of Eq. (16) describes oscillations around the equilibrium position θ=0 with circular frequency ω and amplitude C=(A2+B2)1/2.

FIGURE 9.4.7.

The simple pendulum.

The linear model does not adequately describe the possible motions of the pendulum for large values of θ. For instance, the equilibrium solution θ(t)π of Eq. (15), with the pendulum standing straight up, does not satisfy the linear equation in (16). Nor does Eq. (17) include the situation in which the pendulum “goes over the top” repeatedly, so that θ(t) is a steadily increasing rather than an oscillatory function of t. To investigate these phenomena we must analyze the nonlinear equation θ+ω2 sin θ=0 rather than merely its linearization θ+ω2θ=0. We also want to include the possibility of resistance proportional to velocity, so we consider the general nonlinear pendulum equation

(18)d2θdt2+cdθdt+ω2sinθ=0.

The case c>0 corresponds to damped motion in which there actually is resistance proportional to (angular) velocity. But we examine first the undamped case in which c=0. With x(t)=θ(t) and y(t)=θ(t) the equivalent first-order system is

(19)dxdt=y,dydt=ω2sinx.

We see that this system is almost linear by writing it in the form

(20)dxdt=y,dydt=ω2x+g(x),

where

g(x)=ω2(sinxx)=ω2(x33!x55!+)

has only higher-degree terms.

The critical points of the system in (19) are the points (nπ,0) with n an integer, and its Jacobian matrix is given by

(21)J(x,y)=[01ω2cosx0].

The nature of the critical point (nπ,0) depends on whether n is even or odd.

Even Case: If n=2m is an even integer, then cos nπ=+1, so (21) yields the matrix

J(2mπ,0)=[01ω20]

with characteristic equation λ2+ω2=0 and pure imaginary eigenvalues λ1, λ2=±ωi. The linearization of (19) at (nπ,0) is therefore the system

(22)dudt=v,dvdt=ω2u

for which (0, 0) is the familiar stable center enclosed by elliptical trajectories (as in Example 3 of Section 9.1). Although this is the delicate case for which Theorem 2 of Section 9.2 does not settle the matter, we will see presently that (2mπ,0) is also a stable center for the original nonlinear pendulum system in (19).

Odd Case: If n=2m+1 is an odd integer, then cos nπ=1, so (21) yields the matrix

J((2m+1)π,0)=[01ω20]

with characteristic equation λ2ω2=0 and real eigenvalues λ1, λ2=±ω with different signs. The linearization of (19) at ((2m+1)π,0) is therefore the system

(23)dudt=v,dvdt=ω2u

for which (0, 0) is a saddle point. It follows from Theorem 2 of Section 9.2 that the critical point ((2m+1)π,0) is a similar saddle point for the original nonlinear pendulum system in (19).

The Trajectories: We can see how these “even centers” and “odd saddle points” fit together by solving the system in (19) explicitly for the phase plane trajectories. If we write

dydx=dy/dtdx/dt=ω2sinxy

and separate the variables,

y dy+ω2sinx dx=0,

then integration from x=0 to x=x yields

(24)12y2+ω2(1cosx)=E.

We write E for the arbitrary constant of integration because, if physical units are so chosen that m=L=1, then the first term on the left is the kinetic energy and the second term the potential energy of the mass on the end of the pendulum. Then E is the total mechanical energy; Eq. (24) thus expresses conservation of mechanical energy for the undamped pendulum.

If we solve Eq. (24) for y and use a half-angle identity, we get the equation

(25)y=±2E4ω2sin212x

that defines the phase plane trajectories. Note that the radicand in (25) remains positive if E>2ω2. Figure 9.4.8 shows (along with a direction field) the results of plotting these trajectories for various values of the energy E.

FIGURE 9.4.8.

Position–velocity phase plane portrait for the undamped pendulum system x=y,y=sinx. The (black) separatrices are emphasized.

The emphasized separatrices in Fig. 9.4.8 correspond to the critical value E=2ω2 of the energy; they enter and leave the unstable critical points (nπ,0) with n an odd integer. Following the arrows along a separatrix, the pendulum theoretically approaches a balanced vertical position θ=x=(2m+1)π with just enough energy to reach it but not enough to “go over the top.” The instability of this equilibrium position indicates that this behavior may never be observed in practice!

The simple closed trajectories encircling the stable critical points—all of which correspond to the downward position θ=2mπ of the pendulum—represent periodic oscillations of the pendulum back and forth around the stable equilibrium position θ=0. These correspond to energies E<2ω2 that are insufficient for the pendulum to ascend to the vertical upward position—so its back-and-forth motion is that which we normally associate with a “swinging pendulum.”

The unbounded trajectories with E>2ω2 represent whirling motions of the pendulum in which it goes over the top repeatedly—in a clockwise direction if y(t) remains positive, in a counterclockwise direction if y(t) is negative.

Period of Undamped Oscillation

If the pendulum is released from rest with initial conditions

(26)x(0)=θ(0)=α,y(0)=θ(0)=0,

then Eq. (24) with t=0 reduces to

(27)ω2(1cosα)=E.

Hence E<2ω2 if 0<α<π, so a periodic oscillation of the pendulum ensues. To determine the period of this oscillation, we subtract Eq. (27) from Eq. (24) and write the result (with x=θ and y=dθ/dt) in the form

(28)12(dθdt)2=ω2(cosθcosα).

The period T of time required for one complete oscillation is four times the amount of time required for θ to decrease from θ=α to θ=0, one-fourth of an oscillation. Hence we solve Eq. (28) for dt/dθ and integrate to get

(29)T=4ω20αdθcosθcosα.

To attempt to evaluate this integral we first use the identity cos θ=12sin2(θ/2) and get

T=2ω0αdθk2sin2(θ/2),

where

k=sinα2.

Next, the substitution u=(1/k)sin(θ/2) yields

T=4ω01du(1u2)(1k2u2).

Finally, the substitution u=sin ϕ gives

(30)T=4ω0π/2dϕ1k2sin2ϕ.

The integral in (30) is the elliptic integral of the first kind that is often denoted by F(k,π/2). Whereas elliptic integrals normally cannot be evaluated in closed form, this integral can be approximated numerically as follows. First we use the binomial series

(31)11x=1+n=113(2n1)24(2n)xn

with x=k2 sin2 ϕ<1 to expand the integrand in (30). Then we integrate termwise using the tabulated integral formula

(32)0π/2sin2nϕ dϕ=π213(2n1)24(2n).

The final result is the formula

(33)T=2πω[1+n=1(13(2n1)24(2n))2k2n]=T0[1+(12)2k2+(1324)2k4+(135246)2k6+]

for the period T of the nonlinear pendulum released from rest with initial angle θ(0)=α, in terms of the linearized period T0=2π/ω and k=sin(α/2).

The infinite series within the second pair of brackets in Eq. (33) gives the factor T/T0 by which the nonlinear period T is longer than the linearized period. The table in Fig. 9.4.9, obtained by summing this series numerically, shows how T/T0 increases as α is increased. Thus T is 0.19% greater than T0 if α=10°, whereas T is 18.03% greater than T0 if α=90°. But even a 0.19% discrepancy is significant—the calculation

(0.0019)×3600secondshour×24hoursday×7daysweek1149 (seconds/week)

shows that the linearized model is quite inadequate for a pendulum clock; a discrepancy of 19 min 9 s after only one week is unacceptable.

FIGURE 9.4.9.

Dependence of the period T of a nonlinear pendulum on its initial angle α.

α T/T0
10° 1.0019
20° 1.0077
30° 1.0174
40° 1.0313
50° 1.0498
60° 1.0732
70° 1.1021
80° 1.1375
90° 1.1803

Damped Pendulum Oscillations

Finally, we discuss briefly the damped nonlinear pendulum. The almost linear first-order system equivalent to Eq. (19) is

(34)dxdt=y,dydt=ω2sinxcy,

and again the critical points are of the form (nπ,0) where n is an integer. In Problems 9 through 11 we ask you to verify that

Figure 9.4.10 illustrates the phase plane trajectories for the more interesting underdamped case, c2<4ω2. Other than the physically unattainable separatrix trajectories that enter unstable saddle points, every trajectory eventually is “trapped” by one of the stable spiral points (nπ,0) with n an even integer. What this means is that even if the pendulum starts with enough energy to go over the top, after a certain (finite) number of revolutions it has lost enough energy that thereafter it undergoes damped oscillations around its stable (lower) equilibrium position.

FIGURE 9.4.10.

Position–velocity phase plane portrait for the damped pendulum system x=y,y=sinx14y. The (black) separatrices are emphasized.

9.4 Problems

In Problems 1 through 4, show that the given system is almost linear with (0, 0) as a critical point, and classify this critical point as to type and stability. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your conclusion.

  1. dxdt=1ex+2y, dydt=x4sin y

  2. dxdt=2sin x+sin y, dydt=sin x+2sin y (Fig. 9.4.11)

    FIGURE 9.4.11.

    Trajectories of the system in Problem 2.

  3. dxdt=ex+2y1, dydt=8x+ey1

  4. dxdt=sin xcos y2y, dydt=4x3cos xsin y

Find and classify each of the critical points of the almost linear systems in Problems 5 through 8. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your findings.

  1. dxdt=x+sin y, dydt=2x

  2. dxdt=y, dydt=sin πxy

  3. dxdt=1exy, dydt=2sin x

  4. dxdt=3sin x+y, dydt=sin x+2y

Critical Points for Damped Pendulum

Problems 9 through 11 deal with the damped pendulum system x=y, y=ω2 sin xcy.

  1. Show that if n is an odd integer, then the critical point (nπ,0) is a saddle point for the damped pendulum system.

  2. Show that if n is an even integer and c2>4ω2, then the critical point (nπ,0) is a nodal sink for the damped pendulum system.

  3. Show that if n is an even integer and c2<4ω2, then the critical point (nπ,0) is a spiral sink for the damped pendulum system.

Critical Points for Mass-Spring System

In each of Problems 12 through 16, a second-order equation of the form x+f(x,x)=0, corresponding to a certain mass-and-spring system, is given. Find and classify the critical points of the equivalent first-order system.

  1. x+20x5x3=0: Verify that the critical points resemble those shown in Fig. 9.4.4.

  2. x+2x+20x5x3=0: Verify that the critical points resemble those shown in Fig. 9.4.6.

  3. x8x+2x3=0: Here the linear part of the force is repulsive rather than attractive (as for an ordinary spring). Verify that the critical points resemble those shown in Fig. 9.4.12. Thus there are two stable equilibrium points and three types of periodic oscillations.

  4. x+4xx2=0: Here the force function is nonsymmetric. Verify that the critical points resemble those shown in Fig. 9.4.13.

  5. x+4x5x3+x5=0: The idea here is that terms through the fifth degree in an odd force function have been retained. Verify that the critical points resemble those shown in Fig. 9.4.14.

Critical Points for Physical Systems

In Problems 17 through 20, analyze the critical points of the indicated system, use a computer system to construct an illustrative position–velocity phase plane portrait, and describe the oscillations that occur.

  1. Example 2 in this section illustrates the case of damped vibrations of a soft mass-spring system. Investigate an example of damped vibrations of a hard mass–spring system by using the same parameters as in Example 2, except now with β=54<0.

  2. Example 2 illustrates the case of damped vibrations of a soft mass–spring system with the resistance proportional to the velocity. Investigate an example of resistance proportional to the square of the velocity by using the same parameters as in Example 2, but with resistance term cx|x| instead of cx in Eq. (12).

  3. Now repeat Example 2 with both the alterations corresponding to Problems 17 and 18. That is, take β=54<0 and replace the resistance term in Eq. (12) with cx|x|.

  4. The equations x=y, y=sin x14y|y| model a damped pendulum system as in Eqs. (34) and Fig. 9.4.10. But now the resistance is proportional to the square of the angular velocity of the pendulum. Compare the oscillations with those that occur when the resistance is proportional to the angular velocity itself.

Period of Oscillation

Problems 21 through 26 outline an investigation of the period T of oscillation of a mass on a nonlinear spring with equation of motion

(35)d2xdt2+ϕ(x)=0.

If ϕ(x)=kx with k>0, then the spring actually is linear with period T0=2π/k.

FIGURE 9.4.12.

The phase portrait for Problem 14.

FIGURE 9.4.13.

The phase portrait for Problem 15.

FIGURE 9.4.14.

The phase portrait for Problem 16.

  1. Integrate once (as in Eq. (6)) to derive the energy equation

    (36)12y2+V(x)=E,

    where y=dx/dt and

    (37)V(x)=0xϕ(u) du.
  2. If the mass is released from rest with initial conditions x(0)=x0, y(0)=0 and periodic oscillations ensue, conclude from Eq. (36) that E=V(x0) and that the time T required for one complete oscillation is

    (38)T=420x0duV(x0)V(u).
  3. If ϕ(x)=kxβx3 as in the text, deduce from Eqs. (37) and (38) that

    (39)T=420x0dx(x02u2)(2kβx02βu2).
  4. Substitute u=x0cos ϕ in (39) to show that

    (40)T=2T0π1ϵ0π/2dϕ1μsin2ϕ,

    where T0=2π/k is the linear period,

    (41)ϵ=βkx02,andμ=12ϵ1ϵ.
  5. Finally, use the binomial series in (31) and the integral formula in (32) to evaluate the elliptic integral in (40) and thereby show that the period T of oscillation is given by

    (42)T=T01ϵ(1+14μ+964μ2+25256μ3+).
  6. If ϵ=βx02/k is sufficiently small that ϵ2 is negligible, deduce from Eqs. (41) and (42) that

    (43)TT0(1+38ϵ)=T0(1+3β8kx02).

    It follows that

    • If β>0, so the spring is soft, then T>T0, and increasing x0 increases T, so the larger ovals in Fig. 9.4.4 correspond to smaller frequencies.

    • If β<0, so the spring is hard, then T<T0, and increasing x0 decreases T, so the larger ovals in Fig. 9.4.2 correspond to larger frequencies.

9.4 Application The Rayleigh, van der Pol, and FitzHugh-Nagumo Equations

Here we present a trio of nonlinear differential equations or systems of equations, drawn from the areas of acoustics, electrical engineering, and neuroscience. Each of these models has been fundamental within its field; taken together, they give some indication of the importance of nonlinear equations across a wide variety of applications.

Rayleigh’s Equation

The British mathematical physicist Lord Rayleigh (John William Strutt, 1842–1919) introduced an equation of the form

(1)mx+kx=axb(x)3

to model the oscillations of a clarinet reed. With y=x we get the autonomous system

(2)x=y,y=kx+ayby3m,

whose phase plane portrait is shown in Fig. 9.4.15 (for the case m=k=a=b=1). The outward and inward spiral trajectories converge to a “limit cycle” solution that corresponds to periodic oscillations of the reed. The period T (and hence the frequency) of these oscillations can be measured on a tx-solution curve plotted as in Fig. 9.4.16. This period of oscillation depends only on the parameters m, k, a, and b in Eq. (1) and is independent of the initial conditions (why?).

FIGURE 9.4.15.

Phase plane portrait for the Rayleigh system in (2) with m=k=a=b=1.

FIGURE 9.4.16.

The tx-solution curve with initial conditions x(0)=0.01,x(0)=0.

Choose your own parameters m, k, a, and b (perhaps the least four nonzero digits in your student ID number), and use an available ODE plotting utility to plot trajectories and solution curves as in Figs. 9.4.15 and 9.4.16. Change one of your parameters to see how the amplitude and frequency of the resulting periodic oscillations are altered.

Van der Pol’s Equation

Figure 9.4.17 shows a simple RLC circuit in which the usual (passive) resistance R has been replaced with an active element (such as a vacuum tube or semiconductor) across which the voltage drop V is given by a known function f(I) of the current I. Of course, V=f(I)=IR for a resistor. If we substitute f(I) for IR in the well-known RLC-circuit equation LI+RI+Q/C=0, then differentiation gives the second-order equation

(3)LI+f(I)I+IC=0.

FIGURE 9.4.17.

A simple circuit with an active element.

In a 1924 study of oscillator circuits in early commercial radios, Balthasar van der Pol (1889–1959) assumed the voltage drop to be given by a nonlinear function of the form f(I)=bI3aI, which with Eq. (3) becomes

(4)LI+(3bI2a)I+IC=0.

This equation is closely related to Rayleigh’s equation and has phase portraits resembling Fig. 9.4.15. Indeed, differentiation of the second equation in (2) and the resubstitution x=y yield the equation

(5)my+(3by2a)y+ky=0,

which has the same form as Eq. (4).

If we denote by τ the time variable in Eq. (4) and make the substitutions I=px, t=τ/LC, the result is

d2xdt2+(3bp2x2a)CLdxdt+x=0.

With p=a/(3b) and μ=aC/L, this gives the standard form

(6)x+μ(x21)x+x=0

of van der Pol’s equation.

For every nonnegative value of the parameter μ, the solution of van der Pol’s equation with x(0)=2, x(0)=0 is periodic, and the corresponding phase plane trajectory is a limit cycle to which the other trajectories converge (as in Fig. 9.4.15). It will be instructive for you to solve van der Pol’s equation numerically and to plot this periodic trajectory for a selection of values from μ=0 to μ=1000 or more. With μ=0 it is a circle of radius 2 (why?). Figure 9.4.18 shows the periodic trajectory with μ=1, and Fig. 9.4.19 shows the corresponding x(t) and y(t) solution curves. When μ is large, van der Pol’s equation is quite “stiff” and the periodic trajectory is more eccentric as in Fig. 9.4.20, which was plotted using Matlab’s stiff ODE solver ode15s. The corresponding x(t) and y(t) solution curves in Figs. 9.4.21 and 9.4.22 reveal surprising behavior of these component functions. Each alternates long intervals of very slow change with periods of abrupt change during very short time intervals that correspond to the “quasi-discontinuities” that are visible in Figs. 9.4.21 and 9.4.22. For instance, Fig. 9.4.23 shows that, between t=1614.28 and t=1614.29, the value of y(t) zooms from near zero to over 1300 and back again. Perhaps you can measure the distance between x- or y-intercepts to show that the period of circuit around the cycle in Fig. 9.4.20 is approximately T=1614. Indeed, this calculation and the construction of figures like those shown here may serve as a good test of the robustness of your computer system’s ODE solver.

FIGURE 9.4.18.

The phase plane trajectory of a periodic solution of van der Pol’s equation with μ=1, as well as some trajectories spiraling in and out.

FIGURE 9.4.19.

x(t) and y(t) solution curves defining the periodic solution of van der Pol’s equation with μ=1.

FIGURE 9.4.20.

The phase plane trajectory of the periodic solution of van der Pol’s equation with μ=1000.

FIGURE 9.4.21.

Graph of x(t) with μ=1000.

FIGURE 9.4.22.

Graph of y(t) with μ=1000.

FIGURE 9.4.23.

The upper spike in the graph of y(t).

You might also plot other trajectories for μ=10, 100 or 1000 that (like the trajectories in Fig. 9.4.18) are “attracted” from within and without by the limit cycle. The origin looks like a spiral point in Fig 9.4.18. Indeed, show that (0, 0) is a spiral source for van der Pol’s equation if 0<μ<2 but is a nodal source if μ2.

The FitzHugh-Nagumo Equations

Since the early experiments of Luigi Galvani (1737–1798) in which electrical stimulus caused the leg muscles of dead frogs to twitch, the electrical properties of neurons, the cells that form the building blocks of the nervous system, have been intensively studied. One of the most important of those properties is the action potential, an electrical signal that travels from the body of a neuron down along its axon (Fig. 9.4.24). Action potentials are the units of information of the nervous system; when an action potential reaches the end of the axon, chemicals known as neurotransmitters are released from the axon terminals. These neurotransmitters then find receptors in the dendrites of other nerve cells, causing action potentials in those “target” neurons, and thus propagating the “message.” Because of the great speed with which action potentials traverse the neuron, they provide a mechanism by which signals can be rapidly transmitted through the nervous system.

Action potentials are particularly known for their all-or-none character. If the stimulus received by a target neuron is below a certain threshold, then no action potential is generated. If this threshold is exceeded, however, then the neuron will “fire” an action potential, or perhaps (if the stimulus is sufficiently strong) several action potentials in succession. In this way, the nervous system’s method of electrical signaling resembles the binary code used by computers.

FIGURE 9.4.24.

The structure of a typical neuron.

In the early 1950’s A. F. Huxley (1917–2012) and A. L. Hodgkin (1914–1998) published a landmark series of papers in which they modeled the action potential in the giant axon of the squid as an electrical circuit. The focus of this model was the neuron’s membrane potential, that is, the voltage difference between the inside and outside of the nerve cell. In its resting state, a typical neuron has a negative membrane potential, that is, the inside of the cell is at a lower voltage than the surrounding medium. (We now know that this voltage difference is largely due to “ion pumps” in the cell membrane, which maintain a lower concentration of positively charged sodium ions inside the cell than in the surrounding medium. These pumps require energy, and indeed a significant portion of the body’s metabolic energy is devoted to this task.) Potassium ions also play an important role in neuron electrical activity.

During an action potential, the membrane potential exhibits a characteristic pattern of sudden and rapid changes through positive and negative values as the electrical signal traverses the neuron. A central goal of Hodgkin and Huxley’s work was to explain these changes in terms of the sodium and potassium conductances of the neuron membrane. Conductance—the reciprocal of resistance—is a measure of the permeability of the membrane to charged ions. An increase in the sodium conductance, for example, allows sodium ions to flow more freely across the membrane, from areas of high concentration to low. Hodgkin and Huxley proposed that during an action potential, an electrical stimulus (arising from another neuron “upstream,” for example) causes changes in the neuron membrane’s sodium and potassium conductances. This results in a series of flows of charged ions, and thus electrical currents, across the cell membrane.

The researchers applied some of the basic principles of circuit theory to model these currents. The result was a system of four nonlinear differential equations, whose variables are the neuron membrane potential together with three other quantities related to the membrane’s sodium and potassium conductances. Not only did the predictions of this model show remarkable accord with experimental results, they also helped point the way to subsequent discoveries in neurophysiology. The Hodgkin-Huxley model was a triumph both of experimental technique and theoretical analysis, and remains today the starting point for mathematical modeling of action potentials. Together with John Eccles, Hodgkin and Huxley were awarded the 1963 Nobel Prize in Physiology or Medicine for their work.

Analysis of the Hodgkin-Huxley model can be challenging, however, because its phase space is four-dimensional, making features such as solution curves difficult to visualize. For this reason, in 1961 Richard FitzHugh (1922–2007) proposed a two-dimensional simplification of the Hodgkin-Huxley model, which was subsequently analyzed in electrical circuit terms by J. Nagumo and others. Whereas the FitzHugh-Nagumo equations are not intended to capture the physiological properties of the neuron as directly as the original Hodgkin-Huxley equations do, this simplified model is important because it displays much of the qualitative behavior characteristic of neuron electrical activity, while offering the advantage of being considerably easier to study. FitzHugh’s model actually is a generalization of van der Pol’s equation (6). You can show that introduction of the variable y=1μx+13x3x in van der Pol’s equation (which results in better phase plane analysis than does simply taking y=x) leads to the system

x=μ(y+x13x3),y=1μx,

which FitzHugh generalized by adding terms:

x=μ(y+x13x3+I),y=1μ(xa+by).

Here a and b are constants and I is a function of the time t. Loosely speaking, x(t) behaves in a manner similar to the neuron membrane potential, I(t) is the electrical stimulus applied to the neuron, and y(t) is a composite of the other three variables in the Hodgkin-Huxley model. To simulate neuron electrical activity, we will use FitzHugh’s values a=0.7, b=0.8, and μ=3, while assigning various constant values to the stimulus I(t).

First, with I(t)0 (corresponding to the resting state of the neuron), you can verify that the system (7) has exactly one equilibrium point at roughly x=1.1994 and y=0.6243. What is striking is the way in which the system responds as I becomes nonzero (corresponding to electrical stimulation of the neuron). Figure 9.4.25 shows the solution curves of the system corresponding to the three constant values I(t)0.15, I(t)0.17, and I(t)0.5. The graph on the left gives the phase plane for x and y, whereas the graph on the right shows x(t) as a function of t. All curves begin at time t=0 at the original equilibrium point (1.1994,0.6243) indicated in the graph on the left.

FIGURE 9.4.25.

Solutions of the FitzHugh-Nagumo equations (7) with three constant values of the electrical stimulus I(t), using a=0.7,b=0.8, and μ=3.

When the stimulus I(t) is held at the constant value 0.15, x (which is analogous to the neuron membrane potential) varies slightly from its rest value, but soon finds a new “perturbed” equilibrium at which it then remains (thin blue curves); this mirrors the behavior of the membrane potential when the neuron receives a stimulus insufficiently large to generate an action potential. The response of the system differs dramatically, however, when the constant stimulus is decreased slightly to 0.17; once again x finds a new equilibrium, but only after exhibiting wide swings downward and back upward (dashed curves); this is suggestive of the firing of a single action potential. Finally, when I(t)0.5, x oscillates repeatedly, in a manner reminiscent of both the Rayleigh and van der Pol systems (thick blue curves); this corresponds to repetitive firing of the neuron.

After using your own computer system’s ODE solver to verify these behaviors, you can investigate what happens with other constant nonzero values of the stimulus I(t). Do all such values lead either to oscillatory phase plane solutions or to ones that converge to the same perturbed equilibrium point? Or do certain different values lead to different perturbed equilibrium points? Do all oscillatory solutions correspond to phase plane curves that appear to converge to a single limit cycle?

The FitzHugh-Nagumo system has proven to be a very useful simplification of the Hodgkin-Huxley system, exhibiting a number of characteristics of neuron electrical activity. For a more detailed discussion see the classic work of L. Edelstein-Keshet, Mathematical Models in Biology (Society for Industrial and Applied Mathematics, 2005).

10 Laplace Transform Methods

10.1 Laplace Transforms and Inverse Transforms

In Chapter 5 we saw that linear differential equations with constant coefficients have numerous applications and can be solved systematically. There are common situations, however, in which the alternative methods of this chapter are preferable. For example, recall the differential equations

mx+cx+kx=F(t)andLI+RI+1CI=E(t)

corresponding to a mass–spring–dashpot system and a series RLC circuit, respectively. It often happens in practice that the forcing term, F(t) or E(t), has discontinuities—for example, when the voltage supplied to an electrical circuit is turned off and on periodically. In this case the methods of Chapter 5 can be quite awkward, and the Laplace transform method is more convenient.

The differentiation operator D can be viewed as a transformation which, when applied to the function f(t), yields the new function D{f(t)}=f(t). The Laplace transformation L involves the operation of integration and yields the new function L{f(t)}=F(s) of a new independent variable s. The situation is diagrammed in Fig. 10.1.1. After learning in this section how to compute the Laplace transform F(s) of a function f(t), we will see in Section 10.2 that the Laplace transform converts a differential equation in the unknown function f(t) into an algebraic equation in F(s). Because algebraic equations are generally easier to solve than differential equations, this is one method that simplifies the problem of finding the solution f(t).

FIGURE 10.1.1.

Transformation of a function: L in analogy with D.

Recall that an improper integral over an infinite interval is defined as a limit of integrals over bounded intervals; that is,

(2)ag(t)dt=limbabg(t)dt.

If the limit in (2) exists, then we say that the improper integral converges; otherwise, it diverges or fails to exist. Note that the integrand of the improper integral in (1) contains the parameter s in addition to the variable of integration t. Therefore, when the integral in (1) converges, it converges not merely to a number, but to a function F of s. As in the following examples, it is typical for the improper integral in the definition of L{f(t)} to converge for some values of s and diverge for others.

Example 1

With f(t)1 for t0, the definition of the Laplace transform in (1) gives

L{1}=0estdt=[1sest]0=limb[1sebs+1s],

and therefore

(3)L{1}=1sfors>0.

As in (3), it’s good practice to specify the domain of the Laplace transform—in problems as well as in examples. Also, in this computation we have used the common abbreviation

(4)[g(t)]a=limb[g(t)]ab.

Remark

The limit we computed in Example 1 would not exist if s<0, for then (1/s)ebs would become unbounded as b+. Hence L{1} is defined only for s>0. This is typical of Laplace transforms; the domain of a transform is normally of the form s>a for some number a.

Example 2

With f(t)=eat for t0, we obtain

L{eat}=0esteatdt=0e(sa)tdt=[e(sa)tsa]t=0.

If sa>0, then e(sa)t0 as t+, so it follows that

(5)L{eat}=1safors>a.

Note here that the improper integral giving L{eat} diverges if sa. It is worth noting also that the formula in (5) holds if a is a complex number. For then, with a=α+iβ,

e(sa)t=eiβte(sα)t0

as t+, provided that s>α=Re[a]; recall that eiβt=cos βt+isin βt.

The Laplace transform L{ta} of a power function is most conveniently expressed in terms of the gamma function Γ(x), which is defined for x>0 by the formula

(6)Γ(x)=0ettx1 dt.

For an elementary discussion of Γ(x), see the subsection on the gamma function in Section 11.4, where it is shown that

(7)Γ(1)=1

and that

(8)Γ(x+1)=xΓ(x)

for x>0. It then follows that if n is a positive integer, then

Γ(n+1)=nΓ(n)=n(n1)Γ(n1)=n(n1)(n2)Γ(n2)=n(n1)(n2)2Γ(2)=n(n1)(n2)21Γ(1);

thus

(9)Γ(n+1)=n!

if n is a positive integer. Therefore, the function Γ(x+1), which is defined and continuous for all x>1, agrees with the factorial function for x=n, a positive integer.

Example 3

Suppose that f(t)=ta where a is real and a>1. Then

L{ta}=0esttadt.

If we substitute u=st, t=u/s, and dt=du/s in this integral, we get

(10)L{ta}=1sa+10euuadu=Γ(a+1)sa+1

for all s>0 (so that u=st>0 ). Because Γ(n+1)=n! if n is a nonnegative integer, we see that

(11)L{tn}=n!sn+1fors>0.

For instance,

L{t}=1s2,L{t2}=2s3,andL{t3}=6s4.

As in Problems 1 and 2, these formulas can be derived immediately from the definition, without the use of the gamma function.

Linearity of Transforms

It is not necessary for us to proceed much further in the computation of Laplace transforms directly from the definition. Once we know the Laplace transforms of several functions, we can combine them to obtain transforms of other functions. The reason is that the Laplace transformation is a linear operation.

The proof of Theorem 1 follows immediately from the linearity of the operations of taking limits and of integration:

L{af(t)+bg(t)}=0est[af(t)+bg(t)]dt=limc0cest[af(t)+bg(t)]dt=a(limc0cestf(t)dt)+b(limc0cestg(t)dt)=aL{f(t)}+bL{g(t)}.

Example 4

The computation of L{tn/2} is based on the known special value

(13)Γ(12)=π

of the gamma function. For instance, it follows that

Γ(52)=32Γ(32)=3212Γ(12)=34π,

using the formula Γ(x+1)=xΓ(x) in (9), first with x=32 and then with x=12. Now the formulas in (10) through (12) yield

L{3t2+4t3/2}=32!s3+4Γ(52)s5/2=6s3+3πs5.

Example 5

Recall that coshkt=(ekt+ekt)/2. If k>0, then Theorem 1 and Example 2 together give

L{coshkt}=12L{ekt}+12L{ekt}=12(1sk+1s+k);

that is,

(14)L{coshkt}=ss2k2for s>k>0.

Similarly,

(15)L{sinhkt}=ks2k2for s>k>0.

Because cos kt=(eikt+eikt)/2, the formula in (5) (with a=ik) yields

L{coskt}=12(1sik+1s+ik)=122ss2(ik)2,

and thus

(16)L{coskt}=ss2+k2for s>0.

(The domain follows from s>Re[ik]=0.) Similarly,

(17)L{sinkt}=ks2+k2for s>0.

Example 6

Applying linearity, the formula in (16), and a familiar trigonometric identity, we get

L{3e2t+2sin23t}=L{3e2t+1cos 6t}=3s2+1sss2+36=3s3+144s72s(s2)(s2+36)for s>0.

Inverse Transforms

According to Theorem 3 of this section, no two different functions that are both continuous for all t0 can have the same Laplace transform. Thus if F(s) is the transform of some continuous function f(t), then f(t) is uniquely determined. This observation allows us to make the following definition: If F(s)=L{f(t)}, then we call f(t) the inverse Laplace transform of F(s) and write

(18)f(t)=L1{F(s)}.

Example 7

Using the Laplace transforms derived in Examples 2, 3, and 5 we see that

L1{1s3}=12t2,L1{1s+2}=e2t,L1{2s2+9}=23sin 3t

and so on.

Notation: Functions and Their Transforms. Throughout this chapter we denote functions of t by lowercase letters. The transform of a function will always be denoted by that same letter capitalized. Thus F(s) is the Laplace transform of f(t) and x(t) is the inverse Laplace transform of X(s).

A table of Laplace transforms serves a purpose similar to that of a table of integrals. The table in Fig. 10.1.2 lists the transforms derived in this section; many additional transforms can be derived from these few, using various general properties of the Laplace transformation (which we will discuss in subsequent sections).

Piecewise Continuous Functions

As we remarked at the beginning of this section, we need to be able to handle certain types of discontinuous functions. The function f(t) is said to be piecewise continuous on the bounded interval atb provided that [a, b] can be subdivided into finitely many abutting subintervals in such a way that

  1. f is continuous in the interior of each of these subintervals; and

  2. f(t) has a finite limit as t approaches each endpoint of each subinterval from its interior.

FIGURE 10.1.2.

A short table of Laplace transforms.

f(t) F(s)
1 1s (s>0)
t 1s2 (s>0)
tn (n0) n!sn+1 (s>0)
ta (a>1) Γ(a+1)sa+1 (s>0)
eat 1sa (s>a)
cos k t ss2+k2 (s>0)
sin k t ks2+k2 (s>0)
cosh k t ss2k2 (s>|k|)
sinh k t ks2k2 (s>|k|)
u(ta) eass (s>0)

We say that f is piecewise continuous for t0 if it is piecewise continuous on every bounded subinterval of [0,+). Thus a piecewise continuous function has only simple discontinuities (if any) and only at isolated points. At such points the value of the function experiences a finite jump, as indicated in Fig. 10.1.3. The jump in f(t) at the point c is defined to be f(c+)f(c), where

f(c+)=limϵ0+f(c+ϵ)andf(c)=limϵ0+f(cϵ).

Perhaps the simplest piecewise continuous (but discontinuous) function is the unit step function, whose graph appears in Fig. 10.1.4. It is defined as follows:

(19)u(t)={0for t<0,1for t0.

Because u(t)=1 for t0 and because the Laplace transform involves only the values of a function for t0, we see immediately that

(20)L{u(t)}=1s (s>0).

The graph of the unit step function ua(t)=u(ta) appears in Fig. 10.1.5. Its jump occurs at t=a rather than at t=0; equivalently,

(21)ua(t)=u(ta)={0for t<a,1for ta.

Example 8

Find L{ua(t)} if a>0.

Solution

We begin with the definition of the Laplace transform. We obtain

L{ua(t)}=0estua(t)dt=aestdt=limb[ests]t=ab;

consequently,

(22)L{ua(t)}=eass(s>0, a>0).

FIGURE 10.1.3.

The graph of a piecewise continuous function; the solid dots indicate values of the function at discontinuities.

FIGURE 10.1.4.

The graph of the unit step function.

FIGURE 10.1.5.

The unit step function ua(t) has a jump at t=a.

General Properties of Transforms

It is a familiar fact from calculus that the integral

abg(t) dt

exists if g is piecewise continuous on the bounded interval [a, b]. Hence if f is piecewise continuous for t0, it follows that the integral

0bestf(t) dt

exists for all b<+. But in order for F(s)—the limit of this last integral as b+—to exist, we need some condition to limit the rate of growth of f(t) as t+. The function f is said to be of exponential order as t+ if there exist nonnegative constants M, c, and T such that

(23)|f(t)|Mectfor tT.

Thus a function is of exponential order provided that it grows no more rapidly (as t+) than a constant multiple of some exponential function with a linear exponent. The particular values of M, c, and T are not so important. What is important is that some such values exist so that the condition in (23) is satisfied.

The condition in (23) merely says that f(t)/ect lies between M and M and is therefore bounded in value for t sufficiently large. In particular, this is true (with c=0) if f(t) itself is bounded. Thus every bounded function—such as cos kt or sin kt—is of exponential order.

If p(t) is a polynomial, then the familiar fact that p(t)et0 as t+ implies that (23) holds (for T sufficiently large) with M=c=1. Thus every polynomial function is of exponential order.

For an example of an elementary function that is continuous and therefore bounded on every (finite) interval, but nevertheless is not of exponential order, consider the function f(t)=et2=exp(t2). Whatever the value of c, we see that

limtf(t)ect=limtet2ect=limtet2ct=+

because t2ct+ as t+. Hence the condition in (23) cannot hold for any (finite) value M, so we conclude that the function f(t)=et2 is not of exponential order.

Similarly, because estet2+ as t+, we see that the improper integral 0estet2 dt that would define L{et2} does not exist (for any s), and therefore that the function et2 does not have a Laplace transform. The following theorem guarantees that piecewise functions of exponential order do have Laplace transforms.

Proof:

First we note that we can take T=0 in (23). For by piecewise continuity, |f(t)| is bounded on [0, T]. Increasing M in (23) if necessary, we can therefore assume that |f(t)|M if 0tT. Because ect1 for t0, it then follows that |f(t)|Mect for all t0.

By a standard theorem on convergence of improper integrals—the fact that absolute convergence implies convergence—it suffices for us to prove that the integral

0|estf(t)|dt

exists for s>c. To do this, it suffices in turn to show that the value of the integral

0b|estf(t)|dt

remains bounded as b+. But the fact that |f(t)|Mect for all t0 implies that

0b|estf(t)|dt0b|estMect|dt=M0be(sc)tdtM0e(sc)tdt=Msc

if s>c. This proves Theorem 2.

We have shown, moreover, that

(24)|F(s)|0|estf(t)|dtMsc

if s>c. When we take limits as s+, we get the following result.

The condition in (25) severely limits the functions that can be Laplace transforms. For instance, the function G(s)=s/(s+1) cannot be the Laplace transform of any “reasonable” function because its limit as s+ is 1, not 0. More generally, a rational function—a quotient of two polynomials—can be (and is, as we shall see) a Laplace transform only if the degree of its numerator is less than that of its denominator.

On the other hand, the hypotheses of Theorem 2 are sufficient, but not necessary, conditions for existence of the Laplace transform of f(t). For example, the function f(t)=1/t fails to be piecewise continuous (at t=0 ), but nevertheless (Example 3 with a=12>1) its Laplace transform

L{t1/2}=Γ(12)s1/2=πs

both exists and violates the condition in (24), which would imply that sF(s) remains bounded as s+.

The remainder of this chapter is devoted largely to techniques for solving a differential equation by first finding the Laplace transform of its solution. It is then vital for us to know that this uniquely determines the solution of the differential equation; that is, that the function of s we have found has only one inverse Laplace transform that could be the desired solution. The following theorem is proved in Chapter 6 of Churchill’s Operational Mathematics, 3rd ed. (New York: McGraw-Hill, 1972).

Thus two piecewise continuous functions of exponential order with the same Laplace transform can differ only at their isolated points of discontinuity. This is of no importance in most practical applications, so we may regard inverse Laplace transforms as being essentially unique. In particular, two solutions of a differential equation must both be continuous, and hence must be the same solution if they have the same Laplace transform.

Historical Remark

Laplace transforms have an interesting history. The integral in the definition of the Laplace transform probably appeared first in the work of Euler. It is customary in mathematics to name a technique or theorem for the next person after Euler to discover it (else there would be several hundred different examples of “Euler’s theorem”). In this case, the next person was the French mathematician Pierre Simon de Laplace (1749–1827), who employed such integrals in his work on probability theory. The so-called operational methods for solving differential equations, which are based on Laplace transforms, were not exploited by Laplace. Indeed, they were discovered and popularized by practicing engineers—notably the English electrical engineer Oliver Heaviside (1850–1925). These techniques were successfully and widely applied before they had been rigorously justified, and around the beginning of the twentieth century their validity was the subject of considerable controversy. One reason is that Heaviside blithely assumed the existence of functions whose Laplace transforms contradict the condition that F(s)0 as s0, thereby raising questions as to the meaning and nature of functions in mathematics. (This is reminiscent of the way Leibniz two centuries earlier had obtained correct results in calculus using “infinitely small” real numbers, thereby raising questions as to the nature and role of numbers in mathematics.)

10.1 Problems

Apply the definition in (1) to find directly the Laplace transforms of the functions described (by formula or graph) in Problems 1 through 10.

  1. f(t)=t

  2. f(t)=t2

  3. f(t)=e3t+1

  4. f(t)=cos t

  5. f(t)=sinht

  6. f(t)=sin2 t

  7. FIGURE 10.1.6.

  8. FIGURE 10.1.7.

  9. FIGURE 10.1.8.

  10. FIGURE 10.1.9.

Use the transforms in Fig. 10.1.2 to find the Laplace transforms of the functions in Problems 11 through 22. A preliminary integration by parts may be necessary.

  1. f(t)=t+3t

  2. f(t)=3t5/24t3

  3. f(t)=t2e3t

  4. f(t)=t3/2e10t

  5. f(t)=1+cosh 5t

  6. f(t)=sin 2t+cos 2t

  7. f(t)=cos2 2t

  8. f(t)=sin 3tcos 3t

  9. f(t)=(1+t)3

  10. f(t)=tet

  11. f(t)=tcos 2t

  12. f(t)=sinh2 3t

Use the transforms in Fig. 10.1.2 to find the inverse Laplace transforms of the functions in Problems 23 through 32.

  1. F(s)=3s4

  2. F(s)=s3/2

  3. F(s)=1s2s5/2

  4. F(s)=1s+5

  5. F(s)=3s4

  6. F(s)=3s+1s2+4

  7. F(s)=53ss2+9

  8. F(s)=9+s4s2

  9. F(s)=10s325s2

  10. F(s)=2s1e3s

  11. Derive the transform of f(t)=sin kt by the method used in the text to derive the formula in (16).

  12. Derive the transform of f(t)=sinhkt by the method used in the text to derive the formula in (14).

  13. Use the tabulated integral

    eax cos bx dx=eaxa2+b2(a cos bx+b sin bx)+C

    to obtain L{cos kt} directly from the definition of the Laplace transform.

  14. Show that the function f(t)=sin(et2) is of exponential order as t+ but that its derivative is not.

  15. Given a>0, let f(t)=1 if 0t<a, f(t)=0 if ta. First, sketch the graph of the function f, making clear its value at t=a. Then express f in terms of unit step functions to show that L{f(t)}=s1(1eas).

  16. Given that 0<a<b, let f(t)=1 if at<b, f(t)=0 if either t<a or tb. First, sketch the graph of the function f, making clear its values at t=a and t=b. Then express f in terms of unit step functions to show that L{f(t)}=s1(easebs).

  17. The unit staircase function is defined as follows:

    f(t)=nifn1t<n,n=1, 2, 3,

    (a) Sketch the graph of f to see why its name is appropriate. (b) Show that

    f(t)=n=0u(tn)

    for all t0. (c) Assume that the Laplace transform of the infinite series in part (b) can be taken termwise (it can). Apply the geometric series to obtain the result

    L{f(t)}=1s(1es).
  18. (a) The graph of the function f is shown in Fig. 10.1.10. Show that f can be written in the form

    f(t)=n=0(1)nu(tn).

    (b) Use the method of Problem 39 to show that

    L{f(t)}=1s(1+es).

    FIGURE 10.1.10.

    The graph of the function of Problem 40.

  19. The graph of the square-wave function g(t) is shown in Fig. 10.1.11. Express g in terms of the function f of Problem 40 and hence deduce that

    L{g(t)}=1ess(1+es)=1stanhs2.

    FIGURE 10.1.11.

    The graph of the function of Problem 41.

  20. Given constants a and b, define h(t) for t0 by

    h(t)={aif n1t<n and n is odd;bif n1t<n and n is even.

    Sketch the graph of h and apply one of the preceding problems to show that

    L{h(t)}=a+bess(1+es).

10.1 Application Computer Algebra Transforms and Inverse Transforms

If f(t)=tcos 3t, then the definition of the Laplace transform gives the improper integral

F(s)=L{f(t)}=0testcos 3t dt,

whose evaluation would appear to require a tedious integration by parts. Consequently a computer algebra Laplace transforms package is useful for the quick calculation of transforms. Maple contains the integral transforms package inttrans, and the commands

with(inttrans):
f := t*cos(3*t):
F := laplace(f, t, s);

yield immediately the Laplace transform F(s)=(s29)/(s2+9)2, as do the Mathematica commands

f = t*Cos[3*t];
F = LaplaceTransform[f, t, s]

and the Wolfram|Alpha query

laplace transform t*cos(3t)

We can recover the original function f(t)=tcos 3t with the Maple command

invlaplace(F, s, t);

or the Mathematica command

InverseLaplaceTransform[F, s, t]

or the Wolfram|Alpha query

inverse laplace transform (s^2 - 9)/(s^2 + 9)^2

Remark

Note carefully the order of s and t in the preceding Maple and Mathematica commands—first t, then s when transforming; first s, then t when inverse transforming.

You can use these computer algebra commands to check the answers to Problems 11 through 32 in this section, as well as a few interesting problems of your own selection.

10.2 Transformation of Initial Value Problems

We now discuss the application of Laplace transforms to solve a linear differential equation with constant coefficients, such as

(1)ax(t)+bx(t)+cx(t)=f(t),

with given initial conditions x(0)=x0 and x(0)=x0. By the linearity of the Laplace transformation, we can transform Eq. (1) by separately taking the Laplace transform of each term in the equation. The transformed equation is

(2)aL{x(t)}+bL{x(t)}+cL{x(t)}=L{f(t)};

it involves the transforms of the derivatives x and x of the unknown function x(t). The key to the method is Theorem 1, which tells us how to express the transform of the derivative of a function in terms of the transform of the function itself.

The function f is called piecewise smooth on the bounded interval [a, b] if it is piecewise continuous on [a, b] and differentiable except at finitely many points, with f(t) being piecewise continuous on [a, b]. We may assign arbitrary values to f(t) at the isolated points at which f is not differentiable. We say that f is piecewise smooth for t0 if it is piecewise smooth on every bounded subinterval of [0,+). Figure 10.2.1 indicates how “corners” on the graph of f correspond to discontinuities in its derivative f.

FIGURE 10.2.1.

The discontinuities of f correspond to “corners” on the graph of f.

The main idea of the proof of Theorem 1 is exhibited best by the case in which f(t) is continuous (not merely piecewise continuous) for t0. Then, beginning with the definition of L{f(t)} and integrating by parts, we get

L{f(t)}=0estf(t)dt=[estf(t)]t=0+s0estf(t)dt.

Because of (3), the integrated term estf(t) approaches zero (when s>c) as t+, and its value at the lower limit t=0 contributes f(0) to the evaluation of the preceding expression. The integral that remains is simply L{f(t)}; by Theorem 2 of Section 10.1, the integral converges when s>c. Then L{f(t)} exists when s>c, and its value is that given in Eq. (4). We will defer the case in which f(t) has isolated discontinuities to the end of this section.

Solution of Initial Value Problems

In order to transform Eq. (1), we need the transform of the second derivative as well. If we assume that g(t)=f(t) satisfies the hypotheses of Theorem 1, then that theorem implies that

L{f(t)}=L{g(t)}=sL{g(t)}g(0)=sL{f(t)}f(0)=s[sL{f(t)}f(0)]f(0),

and thus

(5)L{f(t)}=s2F(s)sf(0)f(0).

A repetition of this calculation gives

(6)L{f(t)}=sL{f(t)}f(0)=s3F(s)s2f(0)sf(0)f(0).

After finitely many such steps we obtain the following extension of Theorem 1.

Example 1

Solve the initial value problem

xx6x=0;x(0)=2,x(0)=1.

Solution

With the given initial values, Eqs. (4) and (5) yield

L{x(t)}=sL{x(t)}x(0)=sX(s)2

and

L{x(t)}=s2L{x(t)}sx(0)x(0)=s2X(s)2s+1,

where (according to our convention about notation) X(s) denotes the Laplace transform of the (unknown) function x(t). Hence the transformed equation is

[s2X(s)2s+1][sX(s)2]6[X(s)]=0,

which we quickly simplify to

(s2s6)X(s)2s+3=0.

Thus

X(s)=2s3s2s6=2s3(s3)(s+2).

By the method of partial fractions (of integral calculus), there exist constants A and B such that

2s3(s3)(s+2)=As3+Bs+2,

and multiplication of both sides of this equation by (s3)(s+2) yields the identity

2s3=A(s+2)+B(s3).

If we substitute s=3, we find that A=35; substitution of s=2 shows that B=75. Hence

X(s)=L{x(t)}=35s3+75s+2.

Because L1{1/(sa)}=eat, it follows that

x(t)=35e3t+75e2t

is the solution of the original initial value problem. Note that we did not first find the general solution of the differential equation. The Laplace transform method directly yields the desired particular solution, automatically taking into account—via Theorem 1 and its corollary—the given initial conditions.

Remark

In Example 1 we found the values of the partial-fraction coefficients A and B by the “trick” of separately substituting the roots s=3 and s=2 of the original denominator s2s6=(s3)(s+2) into the equation

2s3=A(s+2)+B(s3)

that resulted from clearing fractions. In lieu of any such shortcut, the “sure-fire” method is to collect coefficients of powers of s on the right-hand side,

2s3=(A+B)s+(2A3).

Then upon equating coefficients of terms of like degree, we get the linear equations

A+B=2,2A3B=3,

which are readily solved for the same values A=35 and B=75.

Example 2

Forced mass-spring system Solve the initial value problem

x+4x=sin 3t;x(0)=x(0)=0.

Such a problem arises in the motion of a mass-and-spring system with external force, as shown in Fig. 10.2.2.

Solution

Because both initial values are zero, Eq. (5) yields L{x(t)}=s2X(s). We read the transform of sin 3t from the table in Fig. 10.1.2 (Section 10.1) and thereby get the transformed equation

s2X(s)+4X(s)=3s2+9.

FIGURE 10.2.2.

A mass–and–spring system satisfying the initial value problem in Example 2. The mass is initially at rest in its equilibrium position.

Therefore,

X(s)=3(s2+4)(s2+9).

The method of partial fractions calls for

3(s2+4)(s2+9)=As+Bs2+4+Cs+Ds2+9.

The sure-fire approach would be to clear fractions by multiplying both sides by the common denominator, and then collect coefficients of powers of s on the right-hand side. Equating coefficients of like powers on the two sides of the resulting equation would then yield four linear equations that we could solve for A, B, C, and D.

However, here we can anticipate that A=C=0, because neither the numerator nor the denominator on the left involves any odd powers of s, whereas nonzero values for A or C would lead to odd-degree terms on the right. So we replace A and C with zero before clearing fractions. The result is the identity

3=B(s2+9)+D(s2+4)=(B+D)s2+(9B+4D).

When we equate coefficients of like powers of s we get the linear equations

B+D=0,9B+4D=3,

which are readily solved for B=35 and D=35. Hence

X(s)=L{x(t)}=3102s2+4153s2+9.

Because L{sin 2t}=2/(s2+4) and L{sin 3t}=3/(s2+9), it follows that

x(t)=310sin 2t15sin 3t.

Figure 10.2.3 shows the graph of this period 2π position function of the mass. Note that the Laplace transform method again gives the solution directly, without the necessity of first finding the complementary function and a particular solution of the original nonhomogeneous differential equation. Thus nonhomogeneous equations are solved in exactly the same manner as are homogeneous equations.

FIGURE 10.2.3.

The position function x(t) in Example 2.

Examples 1 and 2 illustrate the solution procedure that is outlined in Fig. 10.2.4.

FIGURE 10.2.4.

Using the Laplace transform to solve an initial value problem.

Linear Systems

Laplace transforms are used frequently in engineering problems to solve linear systems in which the coefficients are all constants. When initial conditions are specified, the Laplace transform reduces such a linear system of differential equations to a linear system of algebraic equations in which the unknowns are the transforms of the solution functions. As Example 3 illustrates, the technique for a system is essentially the same as for a single linear differential equation with constant coefficients.

Example 3

Dual mass-spring system Solve the system

(8)2x=6x+2y,y=2x2y+40 sin 3t,

subject to the initial conditions

(9)x(0)=x(0)=y(0)=y(0)=0.

Thus the force f(t)=40 sin 3t is applied to the second mass of Fig. 10.2.5, beginning at time t=0 when the system is at rest in its equilibrium position.

FIGURE 10.2.5.

A mass–and–spring system satisfying the initial value problem in Example 3. Both masses are initially at rest in their equilibrium positions.

Solution

We write X(s)=L{x(t)} and Y(s)=L{y(t)}. Then the initial conditions in (9) imply that

L{x(t)}=s2X(s)andL{y(t)}=s2Y(s).

Because L{sin 3t}=3/(s2+9), the transforms of the equations in (8) are the equations

2s2X(s)=6X(s)+2Y(s),s2Y(s)=2X(s)2Y(s)+120s2+9.

Thus the transformed system is

(10)(s2+3)X(s)Y(s)=0,2X(s)+(s2+2)Y(s)=120s2+9.

The determinant of this pair of linear equations in X(s) and Y(s) is

|s2+312s2+2|=(s2+3)(s2+2)2=(s2+1)(s2+4),

and we readily solve—using Cramer’s rule, for instance—the system in (10) for

(11a)X(s)=120(s2+1)(s2+4)(s2+9)=5s2+18s2+4+3s2+9

and

(11b)Y(s)=120(s2+3)(s2+1)(s2+4)(s2+9)=10s2+1+8s2+418s2+9.

The partial fraction decompositions in Eqs. (11a) and (11b) are readily found using the method of Example 2. For instance, noting that the denominator factors are linear in s2, we can write

120(s2+1)(s2+4)(s2+9)=As2+1+Bs2+4+Cs2+9,

and it follows that

(12)120=A(s2+4)(s2+9)+B(s2+1)(s2+9)+C(s2+1)(s2+4).

Substitution of s2=1 (that is, s=i, a zero of the factor s2+1) in Eq. (12) gives 120=A·3·8, so A=5. Similarly, substitution of s2=4 in Eq. (12) yields B=8, and substitution of s2=9 yields C=3. Thus we obtain the partial fraction decomposition shown in Eq. (11a).

At any rate, the inverse Laplace transforms of the expressions in Eqs. (11a) and (11b) give the solution

x(t)=5 sint4 sin 2t+sin 3t,y(t)=10 sint+4 sin 2t6 sin 3t.

Figure 10.2.6 shows the graphs of these two period 2π position functions of the two masses.

FIGURE 10.2.6.

The position functions x(t) and y(t) in Example 3.

The Transform Perspective

Let us regard the general constant-coefficient second-order equation as the equation of motion

mx+cx+kx=f(t)

of the familiar mass–spring–dashpot system (Fig. 10.2.7). Then the transformed equation is

FIGURE 10.2.7.

A mass–spring–dashpot system with external force f(t).

(13)m[s2X(s)sx(0)x(0)]+c[sX(s)x(0)]+kX(s)=F(s).

Note that Eq. (13) is an algebraic equation—indeed, a linear equation—in the “unknown” X(s). This is the source of the power of the Laplace transform method:

Linear differential equations are transformedinto readily solved algebraic equations.

If we solve Eq. (13) for X(s), we get

(14)X(s)=F(s)Z(s)+I(s)Z(s),

where

Z(s)=ms2+cs+kandI(s)=mx(0)s+mx(0)+cx(0).

Note that Z(s) depends only on the physical system itself. Thus Eq. (14) presents X(s)=L{x(t)} as the sum of a term depending only on the external force and one depending only on the initial conditions. In the case of an underdamped system, these two terms are the transforms

L{xsp(t)}=F(s)Z(s)andL{xtr(t)}=I(s)Z(s)

of the steady periodic solution and the transient solution, respectively. The only potential difficulty in finding these solutions is in finding the inverse Laplace transform of the right-hand side in Eq. (14). Much of the remainder of this chapter is devoted to finding Laplace transforms and inverse transforms. In particular, we seek those methods that are sufficiently powerful to enable us to solve problems that—unlike those in Examples 1 and 2—cannot be solved readily by the methods of Chapter 5.

Additional Transform Techniques

Example 4

Show that

L{teat}=1(sa)2.

Solution

If f(t)=teat, then f(0)=0 and f(t)=eat+ateat. Hence Theorem 1 gives

L{eat+ateat}=L{f(t)}=sL{f(t)}=sL{teat}.

It follows from the linearity of the transform that

L{eat}+aL{teat}=sL{teat}.

Hence

(15)L{teat}=L{eat}sa=1(sa)2

because L{eat}=1/(sa).

Example 5

Find L{tsin kt}.

Solution

Let f(t)=tsin kt. Then f(0)=0 and

f(t)=sinkt+ktcoskt.

The derivative involves the new function tcos kt, so we note that f(0)=0 and differentiate again. The result is

f(t)=2k cos ktk2t sin kt.

But L{f(t)}=s2L{f(t)} by the formula in (5) for the transform of the second derivative, and L{cos kt}=s/(s2+k2), so we have

2kss2+k2k2L{tsinkt}=s2L{tsinkt}.

Finally, we solve this equation for

(16)L{tsinkt}=2ks(s2+k2)2.

This procedure is considerably more pleasant than the alternative of evaluating the integral

L{tsinkt}=0testsinktdt.

Examples 4 and 5 exploit the fact that if f(0)=0, then differentiation of f corresponds to multiplication of its transform by s. It is reasonable to expect the inverse operation of integration (antidifferentiation) to correspond to division of the transform by s.

Proof:

Because f is piecewise continuous, the fundamental theorem of calculus implies that

g(t)=0tf(τ) dτ

is continuous and that g(t)=f(t) where f is continuous; thus g is continuous and piecewise smooth for t0. Furthermore,

|g(t)|0t|f(τ)|dτM0tecτdτ=Mc(ect1)<Mcect,

so g(t) is of exponential order as t+. Hence we can apply Theorem 1 to g; this gives

L{f(t)}=L{g(t)}=sL{g(t)}g(0).

Now g(0)=0, so division by s yields

L{0tf(τ)dτ}=L{g(t)}=L{f(t)}s,

which completes the proof.

Example 6

Find the inverse Laplace transform of

G(s)=1s2(sa).

Solution

In effect, Eq. (18) means that we can delete a factor of s from the denominator, find the inverse transform of the resulting simpler expression, and finally integrate from 0 to t (to “correct” for the missing factor s). Thus

L1{1s(sa)}=0tL1{1sa}dτ=0teaτdτ=1a(eat1).

We now repeat the technique to obtain

L1{1s2(sa)}=0tL1{1s(sa)}dτ=0t1a(eaτ1)dτ=[1a(1aeaττ)]0t=1a2(eatat1).

This technique is often a more convenient way than the method of partial fractions for finding an inverse transform of a fraction of the form P(s)/[snQ(s)].

Proof of Theorem 1:

We conclude this section with the proof of Theorem 1 in the general case in which f is merely piecewise continuous. We need to prove that the limit

limb0bestf(t)dt

exists and also need to find its value. With b fixed, let t1,t2,,tk1 be the points interior to the interval [0,b] at which f is discontinuous. Let t0=0 and tk=b. Then we can integrate by parts on each interval (tn1,tn) where f is continuous. This yields

(19)0bestf(t)dt=n=1ktn1tnestf(t)dt=n=1k[estf(t)]tn1tn+n=1kstn1tnestf(t)dt.

Now the first summation

(20)n=1k[estf(t)]tn1tn=[f(t0)+est1f(t1)]+[est1f(t1)+est2f(t2)]++[estk2f(tk2)+estk1f(tk1)]+[estk1f(tk1)+estkf(tk)]

in (19) telescopes down to f(t0)+estkf(tk)=f(0)+esbf(b), and the second summation adds up to s times the integral from t0=0 to tk=b. Therefore (19) reduces to

0bestf(t)dt=f(0)+esbf(b)+s0bestf(t)dt.

But from Eq. (3) we get

|esbf(b)|esbMecb=Meb(sc)0

if s>c. Therefore, finally taking limits (with s fixed) as b+ in the preceding equation, we get the desired result

L{f(t)}=sL{f(t)}f(0).

Extension of Theorem 1

Now suppose that the function f is only piecewise continuous (instead of continuous), and let t1,t2,t3, be the points (for t>0) where either f or f is discontinuous. The fact that f is piecewise continuous includes the assumption that—within each interval [tn1,tn] between successive points of discontinuity—f agrees with a function that is continuous on the whole closed interval and has “endpoint values”

f(tn1+)=limttn1+f(t)andf(tn)=limttnf(t)

that may not agree with the actual values f(tn1) and f(tn). The value of an integral on an interval is not affected by changing the values of the integrand at the endpoints. However, if the fundamental theorem of calculus is applied to find the value of the integral, then the antiderivative function must be continuous on the closed interval. We therefore use the “continuous from within the interval” endpoint values above in evaluating (by parts) the integrals on the right in (19). The result is

(20′)n=1k[estf(t)]tn1tn=[f(t0+)+est1f(t1)]+[est1f(t1+)+est2f(t2)]++[estk2f(tk2+)+estk1f(tk1)]+[estk1f(tk1+)+estkf(tk)]=f(0+)n=1k1jf(tn)+esbf(b),

where

(21)jf(tn)=f(tn+)f(tn)

denotes the (finite) jump in f(t) at t=tn. Assuming that L{f(t)} exists, we therefore get the generalization

(22)L{f(t)}=sF(s)f(0+)n=1estnjf(tn)

of L{f(t)}=sF(s)f(0) when we now take the limit in (19) as b+.

Example 7

Let f(t)=1+t be the unit staircase function; its graph is shown in Fig. 10.2.8. Then f(0)=1, f(t)0, and jf(n)=1 for each integer n=1, 2, 3,. Hence Eq. (22) yields

FIGURE 10.2.8.

The graph of the unit staircase function of Example 7.

0=sF(s)1n=1ens,

so the Laplace transform of f(t) is

F(s)=1sn=0ens=1s(1es).

In the last step we used the formula for the sum of a geometric series,

n=0xn=11x,

with x=es<1.

10.2 Problems

Use Laplace transforms to solve the initial value problems in Problems 1 through 16.

  1. x+4x=0x(0)=5, x(0)=0

  2. x+9x=0x(0)=3, x(0)=4

  3. xx2x=0x(0)=0, x(0)=2

  4. x+8x+15x=0; x(0)=2, x(0)=3

  5. x+x=sin 2tx(0)=0=x(0)

  6. x+4x=cos tx(0)=0=x(0)

  7. x+x=cos 3t; x(0)=1, x(0)=0

  8. x+9x=1x(0)=0=x(0)

  9. x+4x+3x=1; x(0)=0=x(0)

  10. x+3x+2x=t; x(0)=0, x(0)=2

  11. x=2x+y, y=6x+3y; x(0)=1, y(0)=2

  12. x=x+2y, y=x+et; x(0)=y(0)=0

  13. x+2y+x=0, xy+y=0; x(0)=0, y(0)=1

  14. x+2x+4y=0, y+x+2y=0; x(0)=y(0)=0, x(0)=y(0)=1

  15. x+x+y+2xy=0, y+x+y+4x2y=0; x(0)=y(0)=1, x(0)=y(0)=0

  16. x=x+z, y=x+y, z=2xz; x(0)=1, y(0)=0; z(0)=0

Apply Theorem 2 to find the inverse Laplace transforms of the functions in Problems 17 through 24.

  1. F(s)=1s(s3)

  2. F(s)=3s(s+5)

  3. F(s)=1s(s2+4)

  4. F(s)=2s+1s(s2+9)

  5. F(s)=1s2(s2+1)

  6. F(s)=1s(s29)

  7. F(s)=1s2(s21)

  8. F(s)=1s(s+1)(s+2)

  9. Apply Theorem 1 to derive L{sin kt} from the formula for L{cos kt}.

  10. Apply Theorem 1 to derive L{coshkt} from the formula for L{sinhkt}.

    1. Apply Theorem 1 to show that

      L{tneat}=nsaL{tn1eat}.
    2. Deduce that L{tneat}=n!/(sa)n+1 for n=1, 2, 3,.

Apply Theorem 1 as in Example 5 to derive the Laplace transforms in Problems 28 through 30.

  1. L{tcos kt}=s2k2(s2+k2)2

  2. L{tsinhkt}=2ks(s2k2)2

  3. L{tcoshkt}=s2+k2(s2k2)2

  4. Apply the results in Example 5 and Problem 28 to show that

    L1{1(s2+k2)2}=12k3(sinktktcoskt).

Apply the extension of Theorem 1 in Eq. (22) to derive the Laplace transforms given in Problems 32 through 37.

  1. L{u(ta)}=s1eas for a>0.

  2. If f(t)=1 on the interval [a, b] (where 0<a<b) and f(t)=0 otherwise, then

    L{f(t)}=easebss.
  3. If f(t)=(1)t is the square-wave function whose graph is shown in Fig. 10.2.9, then

    L{f(t)}=1stanhs2.

    (Suggestion: Use the geometric series.)

    FIGURE 10.2.9.

    The graph of the square-wave function of Problem 34.

  4. If f(t) is the unit on—off function whose graph is shown in Fig. 10.2.10, then

    L{f(t)}=1s(1+es).

    FIGURE 10.2.10.

    The graph of the on–off function of Problem 35.

  5. If g(t) is the triangular wave function whose graph is shown in Fig. 10.2.11, then

    L{g(t)}=1s2tanhs2.

    FIGURE 10.2.11.

    The graph of the triangular wave function of Problem 36.

  6. If f(t) is the sawtooth function whose graph is shown in Fig. 10.2.12, then

    L{f(t)}=1s2ess(1es).

    (Suggestion: Note that f(t)1 where it is defined.)

    FIGURE 10.2.12.

    The graph of the sawtooth function of Problem 37.

10.2 Application Transforms of Initial Value Problems

The typical computer algebra system knows Theorem 1 and its corollary, hence can transform not only functions (as in the Section 10.1 application), but also entire initial value problems. We illustrate the technique here with Mathematica and in the Section 10.3 application with Maple. Consider the initial value problem

x+4x=sin 3t,x(0)=x(0)=0

of Example 2. First we define the differential equation with its initial conditions, then load the Laplace transform package.

de = x″ [t] + 4* x[t] == Sin[3* t]
inits = {x[0] > 0,x > 0}

The Laplace transform of the differential equation is given by

DE = LaplaceTransform[ de, t, s ]

The result of this command—which we do not show explicitly here—is a linear (algebraic) equation in the as yet unknown LaplaceTransform[x[t],t,s]. We proceed to solve for this transform X(s) of the unknown function x(t) and substitute the initial conditions.

X = Solve[DE, LaplaceTransform[x[t],t,s]]
X = X // Last // Last // Last
X = X /. inits
3(s2+4)(s2+9)

Finally we need only compute an inverse transform to find x(t).

x = InverseLaplaceTransform[X,s,t]
15(3cos(t)sin(t)sin(3t))
x /. {Cos[t] Sin[t] > 1/2 Sin[2t]}// Expand
310sin(2t)15sin(3t)

Of course we could probably get this result immediately with DSolve, but the intermediate output generated by the steps shown here can be quite instructive. You can try it for yourself with the initial value problems in Problems 1 through 16.

10.3 Translation and Partial Fractions

As illustrated by Examples 1 and 2 of Section 10.2, the solution of a linear differential equation with constant coefficients can often be reduced to the matter of finding the inverse Laplace transform of a rational function of the form

(1)R(s)=P(s)Q(s)

where the degree of P(s) is less than that of Q(s). The technique for finding L1{R(s)} is based on the same method of partial fractions that we use in elementary calculus to integrate rational functions. The following two rules describe the partial fraction decomposition of R(s), in terms of the factorization of the denominator Q(s) into linear factors and irreducible quadratic factors corresponding to the real and complex zeros, respectively, of Q(s).

Finding L1{R(s)} involves two steps. First we must find the partial fraction decomposition of R(s), and then we must find the inverse Laplace transform of each of the individual partial fractions of the types that appear in (2) and (3). The latter step is based on the following elementary property of Laplace transforms.

Proof:

If we simply replace s with sa in the definition of F(s)=L{f(t)}, we obtain

F(sa)=0e(sa)tf(t)dt=0est[eatf(t)]dt=L{eatf(t)}.

This is Eq. (4), and it is clear that Eq. (5) is the same.

If we apply the translation theorem to the formulas for the Laplace transforms of tn, cos kt, and sin kt that we already know—multiplying each of these functions by eat and replacing s with sa in the transforms—we get the following additions to the table in Fig. 10.1.2.

f(t) F(s)
eattn n!(sa)n+1 (s>a) (6)
eat cos kt sa(sa)2+k2 (s>a) (7)
eat sin kt k(sa)2+k2 (s>a) (8)

For ready reference, all the Laplace transforms derived in this chapter are listed in the table of transforms that appears in the endpapers.

FIGURE 10.3.1.

The mass–spring–dashpot system of Example 1.

Example 1

Damped mass-spring system Consider a mass-and-spring system with m=12, k=17, and c=3 in mks units (Fig. 10.3.1). As usual, let x(t) denote the displacement of the mass m from its equilibrium position. If the mass is set in motion with x(0)=3 and x(0)=1, find x(t) for the resulting damped free oscillations.

Solution

The differential equation is 12x+3x+17x=0, so we need to solve the initial value problem

x+6x+34x=0;x(0)=3,x(0)=1.

We take the Laplace transform of each term of the differential equation. Because (obviously) L{0}0, we get the equation

[s2X(s)3s1]+6[sX(s)3]+34X(s)=0,

which we solve for

X(s)=3s+19s2+6s+34=3s+3(s+3)2+25+25(s+3)2+25.

Applying the formulas in (7) and (8) with a=3 and k=5, we now see that

x(t)=e3t(3 cos 5t+2 sin 5t).

Figure 10.3.2 shows the graph of this rapidly decaying damped oscillation.

FIGURE 10.3.2.

The position function x(t) in Example 1.

Example 2 illustrates a useful technique for finding the partial fraction coefficients in the case of nonrepeated linear factors.

Example 2

Find the inverse Laplace transform of

R(s)=s2+1s32s28s.

Solution

Note that the denominator of R(s) factors as Q(s)=s(s+2)(s4). Hence

s2+1s32s28s=As+Bs+2+Cs4.

Multiplication of each term of this equation by Q(s) yields

s2+1=A(s+2)(s4)+Bs(s4)+Cs(s+2).

When we successively substitute the three zeros s=0, s=2, and s=4 of the denominator Q(s) in this equation, we get the results

8A=1,12B=5,and24C=17.

Thus A=18, B=512, and C=1724, so

s2+1s32s28s=18s+512s+2+1724s4,

and therefore

L1{s2+1s32s28s}=18+512e2t+1724e4t.

Example 3 illustrates a differentiation technique for finding the partial fraction coefficients in the case of repeated linear factors.

Example 3

Solve the initial value problem

y+4y+4y=t2;y(0)=y(0)=0.

Solution

The transformed equation is

s2Y(s)+4sY(s)+4Y(s)=2s3.

Thus

(9)Y(s)=2s3(s+2)2=As3+Bs2+Cs+D(s+2)2+Es+2.

To find A, B, and C, we multiply both sides by s3 to obtain

(10)2(s+2)2=A+Bs+Cs2+s3F(s),

where F(s)=D(s+2)2+E(s+2)1 is the sum of the two partial fractions corresponding to (s+2)2. Substitution of s=0 in Eq. (10) yields A=12. To find B and C, we differentiate Eq. (10) twice to obtain

(11)4(s+2)3=B+2Cs+3s2F(s)+s3F(s)

and

(12)12(s+2)4=2C+6sF(s)+6s2F(s)+s3F(s).

Now substitution of s=0 in Eq. (11) yields B=12, and substitution of s=0 in Eq. (12) yields C=38.

To find D and E, we multiply each side in Eq. (9) by (s+2)2 to get

(13)2s3=D+E(s+2)+(s+2)2G(s).

where G(s)=As3+Bs2+Cs1, and then differentiate to obtain

(14)6s4=E+2(s+2)G(s)+(s+2)2G(s).

Substitution of s=2 in Eqs. (13) and (14) now yields D=14 and E=38. Thus

Y(s)=12s312s2+38s14(s+2)238s+2,

so the solution of the given initial value problem is

y(t)=14t212t+3814te2t38e2t.

Examples 4, 5, and 6 illustrate techniques for dealing with quadratic factors in partial fraction decompositions.

Example 4

Mass–spring–dashpot system Consider the mass–spring–dashpot system as in Example 1, but with initial conditions x(0)=x(0)=0 and with the imposed external force F(t)=15 sin 2t. Find the resulting transient motion and steady periodic motion of the mass.

Solution

The initial value problem we need to solve is

x+6x+34x=30sin 2t;x(0)=x(0)=0.

The transformed equation is

s2X(s)+6sX(s)+34X(s)=60s2+4.

Hence

X(s)=60(s2+4)[(s+3)2+25]=As+Bs2+4+Cs+D(s+3)2+25.

When we multiply both sides by the common denominator, we get

(15)60=(As+B)[(s+3)2+25]+(Cs+D)(s2+4).

To find A and B, we substitute the zero s=2i of the quadratic factor s2+4 in Eq. (15); the result is

60=(2iA+B)[(2i+3)2+25],

which we simplify to

60=(24A+30B)+(60A+12B)i.

We now equate real parts and imaginary parts on each side of this equation to obtain the two linear equations

24A+30B=60and60A+12B=0,

which are readily solved for A=1029 and B=5029.

To find C and D, we substitute the zero s=3+5i of the quadratic factor (s+3)2+25 in Eq. (15) and get

60=[C(3+5i)+D][(3+5i)2+4],

which we simplify to

60=(186C12D)+(30C30D)i.

Again we equate real parts and imaginary parts; this yields the two linear equations

186C12D=60and30C30D=0,

and we readily find their solution to be C=D=1029.

With these values of the coefficients A, B, C, and D, our partial fractions decomposition of X(s) is

X(s)=129(10s+50s2+4+10s+10(s+3)2+25)=129(10s+252s2+4+10(s+3)45(s+3)2+25).

After we compute the inverse Laplace transforms, we get the position function

x(t)=529(2 cos 2t+5 sin 2t)+229e3t(5 cos 5t2 sin 5t).

The terms of circular frequency 2 constitute the steady periodic forced oscillation of the mass, whereas the exponentially damped terms of circular frequency 5 constitute its transient motion, which disappears very rapidly (see Fig. 10.3.3). Note that the transient motion is nonzero even though both initial conditions are zero.

FIGURE 10.3.3.

The periodic forced oscillation xsp(t), damped transient motion xtr(t), and solution x(t)=xsp(t)+xtr(t) in Example 4.

Resonance and Repeated Quadratic Factors

The following two inverse Laplace transforms are useful in inverting partial fractions that correspond to the case of repeated quadratic factors:

(16)L1{s(s2+k2)2}=12ktsinkt,
(17)L1{s(s2+k2)2}=12k3(sinktktcoskt).

These follow from Example 5 and Problem 31 of Section 10.2, respectively. Because of the presence in Eqs. (16) and (17) of the terms tsin kt and tcos kt, a repeated quadratic factor ordinarily signals the phenomenon of resonance in an undamped mechanical or electrical system.

Example 5

Use Laplace transforms to solve the initial value problem

x+ω02x=F0sinωt;x(0)=0=x(0)

that determines the undamped forced oscillations of a mass on a spring.

Solution

When we transform the differential equation, we get the equation

s2X(s)+ω02X(s)=F0ωs2+ω2,soX(s)=F0ω(s2+ω2)(s2+ω02).

If ωω0, we find without difficulty that

X(s)=F0ωω2ω02(1s2+ω021s2+ω2),

so it follows that

x(t)=F0ωω2ω02(1ω0sinω0t1ωsinωt).

But if ω=ω0, we have

X(s)=F0ω0(s2+ω02)2,

so Eq. (17) yields the resonance solution

(18)x(t)=F02ω02(sinω0tω0tcosω0t).

Remark

The solution curve defined in Eq. (18) bounces back and forth (see Fig. 10.3.4) between the “envelope curves” x=±C(t) that are obtained by writing (18) in the form

x(t)=A(t)cosω0t+B(t)sinω0t

and then defining the usual “amplitude” C=A2+B2. In this case we find that

C(t)=F02ω02ω02t2+1.

This technique for constructing envelope curves of resonance solutions is illustrated further in the application material for this section.

FIGURE 10.3.4.

The resonance solution in (18) with ω0=12 and F0=1, together with its envelope curves x=±C(t).

Example 6

Solve the initial value problem

y(4)+2y+y=4tet;y(0)=y(0)=y(0)=y(3)(0)=0.

Solution

First we observe that

L{y(t)}=s2Y(s),L{y(4)(t)}=s4Y(s),andL{tet}=1(s1)2.

Hence the transformed equation is

(s4+2s2+1)Y(s)=4(s1)2.

Thus our problem is to find the inverse transform of

(19)Y(s)=4(s1)2(s2+1)2=A(s1)2+Bs1+Cs+D(s2+1)2+Es+Fs2+1.

If we multiply by the common denominator (s1)2(s2+1)2, we get the equation

(20)A(s2+1)2+B(s1)(s2+1)2+Cs(s1)2+D(s1)2+Es(s1)2(s2+1)+F(s1)2(s2+1)=4.

Upon substituting s=1 we find that A=1.

Equation (20) is an identity that holds for all values of s. To find the values of the remaining coefficients, we substitute in succession the values s=0, s=1, s=2, s=2, and s=3 in Eq. (20). This yields the system

(21)B+D+F=3,8B4C+4D8E+8F=0,25B+2C+D+10E+5F=21,75B18C+9D90E+45F=21,200B+12C+4D+120E+40F=96

of five linear equations in B, C, D, E, and F. With the aid of a calculator programmed to solve linear systems, we find that B=2, C=2, D=0, E=2, and F=1.

We now substitute in Eq. (19) the coefficients we have found, and thus obtain

Y(s)=1(s1)22s1+2s(s2+1)2+2s+1s2+1.

Recalling Eq. (16), the translation property, and the familiar transforms of cos t and sin t, we see finally that the solution of the given initial value problem is

y(t)=(t2)et+(t+1) sin t+2 cos t.

10.3 Problems

Apply the translation theorem to find the Laplace transforms of the functions in Problems 1 through 4.

  1. f(t)=t4eπt

  2. f(t)=t3/2e4t

  3. f(t)=e2tsin 3πt

  4. f(t)=et/2cos 2(t18π)

Apply the translation theorem to find the inverse Laplace transforms of the functions in Problems 5 through 10.

  1. F(s)=32s4

  2. F(s)=s1(s+1)3

  3. F(s)=1s2+4s+4

  4. F(s)=s+2s2+4s+5

  5. F(s)=3s+5s26s+25

  6. F(s)=2s39s212s+20

Use partial fractions to find the inverse Laplace transforms of the functions in Problems 11 through 22.

  1. F(s)=1s24

  2. F(s)=5s6s23s

  3. F(s)=52ss2+7s+10

  4. F(s)=5s4s3s22s

  5. F(s)=1s35s2

  6. F(s)=1(s2+s6)2

  7. F(s)=1s416

  8. F(s)=s3(s4)4

  9. F(s)=s22ss4+5s2+4

  10. F(s)=1s4+8s2+16

  11. F(s)=s2+3(s2+2s+2)2

  12. F(s)=2s3s2(4s24s+5)2

Use the factorization

s4+4a4=(s22as+2a2)(s2+2as+2a2)

to derive the inverse Laplace transforms listed in Problems 23 through 26.

  1. L1{s3s4+4a4}=cosh atcos at

  2. L1{ss4+4a4}=12a2sinhatsinat

  3. L1{s2s4+4a4}=12a(cosh atsin at+sinh atcos at)

  4. L1{1s4+4a4}=14a3(cosh atsin atsinh atcos at)

Use Laplace transforms to solve the initial value problems in Problems 27 through 38.

  1. x+6x+25x=0; x(0)=2, x(0)=3

  2. x6x+8x=2; x(0)=x(0)=0

  3. x4x=3t; x(0)=x(0)=0

  4. x+4x+8x=et; x(0)=x(0)=0

  5. x(3)+x6x=0; x(0)=0, x(0)=x(0)=1

  6. x(4)x=0; x(0)=1, x(0)=x(0)=x(3)(0)=0

  7. x(4)+x=0; x(0)=x(0)=x(0)=0, x(3)(0)=1

  8. x(4)+13x+36x=0; x(0)=x(0)=0, x(0)=2, x(3)(0)=13

  9. x(4)+8x+16x=0; x(0)=x(0)=x(0)=0, x(3)(0)=1

  10. x(4)+2x+x=e2t; x(0)=x(0)=x(0)=x(3)(0)=0

  11. x+4x+13x=tet; x(0)=0, x(0)=2

  12. x+6x+18x=cos 2t; x(0)=1, x(0)=1

Resonance

Problems 39 and 40 illustrate two types of resonance in a mass–spring–dashpot system with given external force F(t) and with the initial conditions x(0)=x(0)=0.

  1. Suppose that m=1, k=9, c=0, and F(t)=6 cos 3t. Use the inverse transform given in Eq. (16) to derive the solution x(t)=tsin 3t. Construct a figure that illustrates the resonance that occurs.

  2. Suppose that m=1, k=9.04, c=0.4, and F(t)=6et/5 cos 3t. Derive the solution

    x(t)=tet/5sin 3t.

    Show that the maximum value of the amplitude function A(t)=tet/5 is A(5)=5/e. Thus (as indicated in Fig. 10.3.5) the oscillations of the mass increase in amplitude during the first 5 s before being damped out as t+.

    FIGURE 10.3.5.

    The graph of the damped oscillation in Problem 40.

10.3 Application Damping and Resonance Investigations

Here we outline a Maple investigation of the behavior of the mass–spring–dashpot system

(1)mx+cx+kx=F(t), x(0)=x(0)=0

with parameter values

m := 25; c := 10; k := 226;

in response to a variety of possible external forces:

  1. F(t)226

This should give damped oscillations “leveling off” to a constant solution (why?).

  1. F(t)=901 cos 3t

With this periodic external force you should see a steady periodic oscillation with an exponentially damped transient motion (as illustrated in Fig. 5.6.13).

  1. F(t)=900et/5 cos 3t

Now the periodic external force is exponentially damped, and the transform X(s) includes a repeated quadratic factor that signals the presence of a resonance phenomenon. The response x(t) isa constant multiple of that shown in Fig. 10.3.5.

  1. F(t)=900tet/5 cos 3t

We have inserted the factor t to make it a bit more interesting. The solution in this case is illustrated below.

  1. F(t)=162t3et/5 cos 3t

In this case you’ll find that the transform X(s) involves the fifth power of a quadratic factor, and its inverse transform by manual methods would be impossibly tedious.

To illustrate the Maple approach, we first set up the differential equation corresponding to Case 4.

F := 900*t*exp(t/5) *cos(3* t);
de := m* diff(x(t),t$2) + c* diff(x(t),t) + k* x(t) = F;

Then we apply the Laplace transform and substitute the initial conditions.

with(inttrans):
DE := laplace(de, t, s):
X(s) := solve(DE, laplace(x(t), t, s)):
X(s) := simplify(subs(x(0)=0, D(x)(0)=0, X(s)));

At this point the command factor(denom(X(s))) shows that

X(s)=22500(25s2+10s224)(25s2+10s+226)3.

The cubed quadratic factor would be difficult to handle manually, but the command

x(t) := invlaplace(X(s), s, t);

soon yields

x(t)=et/5(tcos 3t+(3t213)sin 3t).

The amplitude function for these damped oscillations is defined by

C(t) := exp(-t/5)*sqrt(t^2 + (3*t^2 1/3)^2);

and finally the command

plot({x(t), C(t), -C(t)}, t=0..40);

produces the plot shown in Fig. 10.3.6. The resonance resulting from the repeated quadratic factor consists of a temporary buildup before the oscillations are damped out.

FIGURE 10.3.6.

The resonance solution and its envelope curves in Case 4.

For a similar solution in one of the other cases listed previously, you need only enter the appropriate force F in the initial command above and then re-execute the subsequent commands. To see the advantage of using Laplace transforms, set up the differential equation de for Case 5 and examine the result of the command

dsolve({de, x(0)=0, D(x)(0)=0}, x(t));

Of course you can substitute your own favorite mass–spring–dashpot parameters for those used here. But it will simplify the calculations if you choose m, c, and k so that

(2)mr2+cr+k=(pr+a)2+b2

where p, a, and b are integers. One way is to select the latter integers first, then use Eq. (2) to determine m, c, and k.

10.4 Derivatives, Integrals, and Products of Transforms

The Laplace transform of the (initially unknown) solution of a differential equation is sometimes recognizable as the product of the transforms of two known functions. For example, when we transform the initial value problem

x+x=cost;x(0)=x(0)=0,

we get

X(s)=s(s2+1)2=ss2+11s2+1=L{cos t}L{sin t}.

This strongly suggests that there ought to be a way of combining the two functions sin t and cos t to obtain a function x(t) whose transform is the product of their transforms. But obviously x(t) is not simply the product of cos t and sin t, because

L{costsint}=L{12sin 2t}=1s2+4s(s2+1)2.

Thus L{cos tsin t}L{cos t}·L{sin t}.

Theorem 1 of this section will tell us that the function

(1)h(t)=0tf(τ)g(tτ)dτ

has the desired property that

(2)L{h(t)}=H(s)=F(s)G(s)

The new function of t defined as the integral in (1) depends only on f and g and is called the convolution of f and g. It is denoted by fg, the idea being that it is a new type of product of f and g, so tailored that its transform is the product of the transforms of f and g.

We will also write f(t)g(t) when convenient. In terms of the convolution product, Theorem 1 of this section says that

L{fg}=L{f}L{g}.

If we make the substitution u=tτ in the integral in (3), we see that

f(t)g(t)=0tf(τ)g(tτ)dτ=t0f(tu)g(u)(du)=0tg(u)f(tu)du=g(t)f(t).

Thus the convolution is commutative: fg=gf.

Example 1

The convolution of cos t and sin t is

(cost)(sint)=0tcosτsin(tτ)dτ.

We apply the trigonometric identity

cosAsinB=12[sin(A+B)sin(AB)]

to obtain

(cost)(sint)=0t12[sintsin(2τt)]dτ=12[τsint+12cos(2τt)]τ=0t;

that is,

(cost)(sint)=12tsint.

And we recall from Example 5 of Section 10.2 that the Laplace transform of 12tsin t is indeed s/(s2+1)2.

Theorem 1 is proved at the end of this section.

Thus we can find the inverse transform of the product F(s)·G(s), provided that we can evaluate the integral

(5′)L1{F(s)G(s)}=0tf(τ)g(tτ)dτ.

Example 2 illustrates the fact that convolution often provides a convenient alternative to the use of partial fractions for finding inverse transforms.

Example 2

With f(t)=sin 2t and g(t)=et, convolution yields

L1{2(s1)(s2+4)}=(sin 2t)et=0tetτsin 2τdτ=et0teτsin 2τdτ=et[eτ5(sin 2τ2 cos 2τ)]0t,

so

L1{2(s1)(s2+4)}=25et15sin 2t25cos 2t.

Differentiation of Transforms

According to Theorem 1 of Section 10.2, if f(0)=0 then differentiation of f(t) corresponds to multiplication of its transform by s. Theorem 2, proved at the end of this section, tells us that differentiation of the transform F(s) corresponds to multiplication of the original function f(t) by t.

Example 3

Find L{t2 sin kt}.

Solution

Equation (8) gives

(9)L{t2sinkt}=(1)2d2ds2(ks2+k2)=dds[2ks(s2+k2)2]=6ks22k3(s2+k2)3.

The form of the differentiation property in Eq. (7) is often helpful in finding an inverse transform when the derivative of the transform is easier to work with than the transform itself.

Example 4

Find L1{tan1(1/s)}.

Solution

The derivative of tan1(1/s) is a simple rational function, so we apply Eq. (7):

L1{tan11s}=1tL1{ddstan11s}=1tL1{1/s21+(1/s)2}=1tL1{1s2+1}=1t(sint).

Therefore,

L1{tan11s}=sintt.

Equation (8) can be applied to transform a linear differential equation having polynomial, rather than constant, coefficients. The result will be a differential equation involving the transform; whether this procedure leads to success depends, of course, on whether we can solve the new equation more readily than the old one.

Example 5

Let x(t) be the solution of Bessel’s equation of order zero,

tx+x+tx=0,

such that x(0)=1 and x(0)=0. This solution of Bessel’s equation is customarily denoted by J0(t). Because

L{x(t)}=sX(s)1andL{x(t)}=s2X(s)s,

and because x and x are each multiplied by t, application of Eq. (6) yields the transformed equation

dds[s2X(s)s]+[sX(s)1]dds[X(s)]=0.

The result of differentiation and simplification is the differential equation

(s2+1)X(s)+sX(s)=0.

This equation is separable—

X(s)X(s)=ss2+1;

its general solution is

X(s)=Cs2+1.

In Problem 39 we outline the argument that C=1. Because X(s)=L{J0(t)}, it follows that

(10)L{J0(t)}=1s2+1.

Integration of Transforms

Differentiation of F(s) corresponds to multiplication of f(t) by t (together with a change of sign). It is therefore natural to expect that integration of F(s) will correspond to division of f(t) by t. Theorem 3, proved at the end of this section, confirms this, provided that the resulting quotient f(t)/t remains well behaved as t0 from the right; that is, provided that

(11)limt0+f(t)t exists and is finite.

Example 6

Find L{(sinht)/t}.

Solution

We first verify that the condition in (11) holds:

limt0sinhtt=limt0etet2t=limt0et+et2=1,

with the aid of l’Hôpital’s rule. Then Eq. (12), with f(t)=sinht, yields

L{sinhtt}=sL{sinht}dσ=sdσσ21=12s(1σ11σ+1)dσ=12[lnσ1σ+1]s.

Therefore,

L{sinhtt}=12lns+1s1,

because ln 1=0.

The form of the integration property in Eq. (13) is often helpful in finding an inverse transform when the indefinite integral of the transform is easier to handle than the transform itself.

Example 7

Find L1{2s/(s21)2}.

Solution

We could use partial fractions, but it is much simpler to apply Eq. (13). This gives

L1{2s(s21)2}=tL1{s2σ(σ21)2dσ}=tL1{[1σ21]s}=tL1{1s21},

and therefore

L1{2s(s21)2}=tsinht.

* Proofs of Theorems

Proof of Theorem 1:

The transforms F(s) and G(s) exist when s>c by Theorem 2 of Section 10.1. For any τ>0 the definition of the Laplace transform gives

G(s)=0esug(u)du=τes(tτ)g(tτ)dt(u=tτ),

and therefore

G(s)=esτ0estg(tτ)dt,

because we may define f(t) and g(t) to be zero for t<0. Then

F(s)G(s)=G(s)0esτf(τ)dτ=0esτf(τ)G(s)dτ=0esτf(τ)(esτ0esτg(tτ)dt)dτ=0(0estf(τ)g(tτ)dt)dτ.

Now our hypotheses on f and g imply that the order of integration may be reversed. (The proof of this requires a discussion of uniform convergence of improper integrals, and can be found in Chapter 2 of Churchill’s Operational Mathematics, 3rd ed. (New York: McGraw-Hill, 1972).) Hence

F(s)G(s)=0(0estf(τ)g(tτ)dτ)dt=0est(0tf(τ)g(tτ)dτ)dt=0est[f(t)g(t)]dt,

and therefore,

F(s)G(s)=L{f(t)g(t)}.

We replace the upper limit of the inner integral with t because g(tτ)=0 whenever τ>t. This completes the proof of Theorem 1.

Proof of Theorem 2:

Because

F(s)=0estf(t)dt,

differentiation under the integral sign yields

F(s)=dds0estf(t)dt=0dds[estf(t)] dt=0est[tf(t)] dt;

thus

F(s)=L{tf(t)},

which is Eq. (6). We obtain Eq. (7) by applying L1 and then dividing by t. The validity of differentiation under the integral sign depends on uniform convergence of the resulting integral; this is discussed in Chapter 2 of the book by Churchill just mentioned.

Proof of Theorem 3:

By definition,

F(σ)=0eσtf(t) dt.

So integration of F(σ) from s to + gives

sF(σ) dσ=s(0eσtf(t) dt)dσ.

Under the hypotheses of the theorem, the order of integration may be reversed (see Churchill’s book once again); it follows that

sF(σ)dσ=0(seσtf(t)dσ)dt=0[eσtt]σ=sf(t)dt=0estf(t)tdt=L{f(t)t}.

This verifies Eq. (12), and Eq. (13) follows upon first applying L1 and then multiplying by t.

10.4 Problems

Find the convolution f(t)g(t) in Problems 1 through 6.

  1. f(t)=t, g(t)1

  2. f(t)=t, g(t)=eat

  3. f(t)=t, g(t)=sint

  4. f(t)=t2, g(t)=cost

  5. f(t)=g(t)=eat

  6. f(t)=eat, g(t)=ebt(ab)

Apply the convolution theorem to find the inverse Laplace transforms of the functions in Problems 7 through 14.

  1. F(s)=1s(s3)

  2. F(s)=1s(s2+4)

  3. F(s)=1(s2+9)2

  4. F(s)=1s2(s2+k2)

  5. F(s)=s2(s2+4)2

  6. F(s)=1s(s2+4s+5)

  7. F(s)=s(s3)(s2+1)

  8. F(s)=ss4+5s2+4

In Problems 15 through 22, apply either Theorem 2 or Theorem 3 to find the Laplace transform of f(t).

  1. f(t)=tsin 3t

  2. f(t)=t2cos 2t

  3. f(t)=te2tcos 3t

  4. f(t)=tetsin2t

  5. f(t)=sintt

  6. f(t)=1cos 2tt

  7. f(t)=e3t1t

  8. f(t)=etett

Find the inverse transforms of the functions in Problems 23 through 28.

  1. F(s)=lns2s+2

  2. F(s)=lns2+1s2+4

  3. F(s)=lns2+1(s+2)(s3)

  4. F(s)=tan13s+2

  5. F(s)=ln(1+1s2)

  6. F(s)=s(s2+1)3

In Problems 29 through 34, transform the given differential equation to find a nontrivial solution such that x(0)=0.

  1. tx+(t2)x+x=0

  2. tx+(3t1)x+3x=0

  3. tx(4t+1)x+2(2t+1)x=0

  4. tx+2(t1)x2x=0

  5. tx2x+tx=0

  6. tx+(4t2)x+(13t4)x=0

  7. Apply the convolution theorem to show that

    L1{1(s1)s}=2etπ0teu2 du=eterft.

    (Suggestion: Substitute u=t.)

In Problems 36 through 38, apply the convolution theorem to derive the indicated solution x(t) of the given differential equation with initial conditions x(0)=x(0)=0.

  1. x+4x=f(t)x(t)=120tf(tτ)sin 2τ dτ

  2. x+2x+x=f(t)x(t)=0tτeτf(tτ) dτ

  3. x+4x+13x=f(t);x(t)=130tf(tτ)e2τ sin 3τ dτ

Termwise Inverse Transformation of Series

In Chapter 2 of Churchill’s Operational Mathematics, the following theorem is proved. Suppose that f(t) is continuous for t0, that f(t) is of exponential order as t+, and that

F(s)=n=0ansn+k+1

where 0k<1 and the series converges absolutely for s>c. Then

f(t)=n=0antn+kΓ(n+k+1).
Apply this result in Problems 39 through 41.
  1. In Example 5 it was shown that

    L{J0(t)}=Cs2+1=Cs(1+1s2)1/2.

    Expand with the aid of the binomial series and then compute the inverse transformation term by term to obtain

    J0(t)=Cn=0(1)nt2n22n(n!)2.

    Finally, note that J0(0)=1 implies that C=1.

  2. Expand the function F(s)=s1/2e1/s in powers of s1 to show that

    L1{1se1/s}=1πtcos 2t.
  3. Show that

    L1{1se1/s}=J0(2t).

10.5 Periodic and Piecewise Continuous Input Functions

Mathematical models of mechanical or electrical systems often involve functions with discontinuities corresponding to external forces that are turned abruptly on or off. One such simple on–off function is the unit step function that we introduced in Section 10.1. Recall that the unit step function at t=a is defined by

(1)ua(t)=u(ta)={0if t<a,1if ta.

The notation ua(t) indicates succinctly where the unit upward step in value takes place (Fig. 10.5.1), whereas u(ta) connotes the sometimes useful idea of a “time delay” a before the step is made.

In Example 8 of Section 10.1 we saw that if a0, then

FIGURE 10.5.1.

The graph of the unit step function at t=a.

(2)L{u(ta)}=eass.

Because L{u(t)}=1/s, Eq. (2) implies that multiplication of the transform of u(t) by eas corresponds to the translation tta in the original independent variable. Theorem 1 tells us that this fact, when properly interpreted, is a general property of the Laplace transformation.

Note that

(4)u(ta)f(ta)={0if t<a,f(ta)if ta.

Thus Theorem 1 implies that L1{easF(s)} is the function whose graph for ta is the translation by a units to the right of the graph of f(t) for t0. Note that the part (if any) of the graph of f(t) to the left of t=0 is “cut off” and is not translated (Fig. 10.5.2). In some applications the function f(t) describes an incoming signal that starts arriving at time t=0. Then u(ta)f(ta) denotes a signal of the same “shape” but with a time delay of a, so it does not start arriving until time t=a.

FIGURE 10.5.2.

Translation of f(t) a units to the right.

Proof of Theorem 1:

From the definition of L{f(t)}, we get

easF(s)=eas0esτf(τ)dτ=0es(τ+a)f(τ)dτ.

The substitution t=τ+a then yields

easF(s)=aestf(ta)dt.

From Eq. (4) we see that this is the same as

easF(s)=0estu(ta)f(ta)dt=L{u(ta)f(ta)},

because u(ta)f(ta)=0 for t<a. This completes the proof of Theorem 1.

Example 1

With f(t)=12t2, Theorem 1 gives

L1{eass3}=u(ta)12(ta)2={0if t<a,12(ta)2if ta(Fig. 10.5.3).

Example 2

Find L{g(t)} if

g(t)={0if t<3,t2if t3(Fig. 10.5.4).

Solution

Before applying Theorem 1, we must first write g(t) in the form u(t3)f(t3). The function f(t) whose translation 3 units to the right agrees (for t3) with g(t)=t2 is f(t)=(t+3)2 because f(t3)=t2. But then

F(s)=L{t2+6t+9}=2s3+6s2+9s,

so now Theorem 1 yields

L{g(t)}=L{u(t3)f(t3)}=e3sF(s)=e3s(2s3+6s2+9s).

FIGURE 10.5.3.

The graph of the inverse transform of Example 1.

FIGURE 10.5.4.

The graph of the function g(t) of Example 2.

FIGURE 10.5.5.

The function f(t) of Examples 3 and 4.

Example 3

Find L{f(t)} if

f(t)={cos 2tif 0t<2π,0if t2π  (Fig.10.5.5).

Solution

We note first that

f(t)=[1u(t2π)]cos 2t=cos 2tu(t2π)cos 2(t2π)

because of the periodicity of the cosine function. Hence Theorem 1 gives

L{f(t)}=L{cos 2t}e2πsL{cos 2t}=s(1e2πs)s2+4.

Example 4

Discontinuous forcing A mass that weighs 32 lb (mass m=1 slug) is attached to the free end of a long light spring that is stretched 1 ft by a force of 4 lb (k=4 lb/ft). The mass is initially at rest in its equilibrium position. Beginning at time t=0 (seconds), an external force F(t)=cos 2t is applied to the mass, but at time t=2π this force is turned off (abruptly discontinued) and the mass is allowed to continue its motion unimpeded. Find the resulting position function x(t) of the mass.

Solution

We need to solve the initial value problem

x+4x=f(t);x(0)=x(0)=0,

where f(t) is the function of Example 3. The transformed equation is

(s2+4)X(s)=F(s)=s(1e2πs)s2+4,

so

X(s)=s(s2+4)2e2πss(s2+4)2.

Because

L1{s(s2+4)2}=14tsin 2t

by Eq. (16) of Section 10.3, it follows from Theorem 1 that

x(t)=14tsin 2tu(t2π)14(t2π)sin 2(t2π)=14[tu(t2π)(t2π)]sin 2t.

If we separate the cases t<2π and t2π, we find that the position function may be written in the form

x(t)={14tsin 2tif t<2π,12πsin 2tif t2π.

As indicated by the graph of x(t) shown in Fig. 10.5.6, the mass oscillates with circular frequency ω=2 and with linearly increasing amplitude until the force is removed at time t=2π. Thereafter, the mass continues to oscillate with the same frequency but with constant amplitude π/2. The force F(t)=cos 2t would produce pure resonance if continued indefinitely, but we see that its effect ceases immediately at the moment it is turned off.

FIGURE 10.5.6.

The graph of the function x(t) of Example 4.

If we were to attack Example 4 with the methods of Chapter 5, we would need to solve one problem for the interval 0t<2π and then solve a new problem with different initial conditions for the interval t2π. In such a situation the Laplace transform method enjoys the distinct advantage of not requiring the solution of different problems on different intervals.

Transforms of Periodic Functions

Periodic forcing functions in practical mechanical or electrical systems often are more complicated than pure sines or cosines. The nonconstant function f(t) defined for t0 is said to be periodic if there is a number p>0 such that

(5)f(t+p)=f(t)

for all t0. The least positive value of p (if any) for which Eq. (11) holds is called the period of f. Such a function is shown in Fig. 10.5.7. Theorem 2 simplifies the computation of theLaplace transform of a periodic function.

FIGURE 10.5.7.

The graph of a function with period p.

Proof:

The definition of the Laplace transform gives

F(s)=0estf(t)dt=n=0np(n+1)pestf(t)dt.

The substitution t=τ+np in the nth integral following the summation sign yields

np(n+1)pestf(t)dt=0pes(τ+np)f(τ+np)dτ=enps0pesτf(τ)dτ

because f(τ+np)=f(τ) by periodicity. Thus

F(s)=n=0(enps0pesτf(τ)dτ)=(1+eps+e2ps+)0pesτf(τ)dτ.

Consequently,

F(s)=11eps0pesτf(τ)dτ.

We use the geometric series

11x=1+x+x2+x3+,

with x=eps<1 (for s>0) to sum the series in the final step. Thus we have derived Eq. (6).

The principal advantage of Theorem 2 is that it enables us to find the Laplace transform of a periodic function without the necessity of an explicit evaluation of an improper integral.

Example 5

Figure 10.5.8 shows the graph of the square wave function f(t)=(1)t/a of period p=2ax denotes the greatest integer not exceeding x. By Theorem 2 theLaplace transform of f(t) is

F(s)=11e2as02aestf(t)dt=11e2as(0aestdt+a2a(1)estdt)=11e2as([1sest]0a[1sest]a2a)=(1eas)2s(1e2as)=1eass(1+eas).

Therefore,

(7a)F(s)=1eass(1+eas)
(7b)=eas/2eas/2s(eas/2+eas/2)=1stanhas2.

FIGURE 10.5.8.

The graph of the square wave function of Example 5.

Example 6

Figure 10.5.9 shows the graph of a triangular wave function g(t) of period p=2a. Becausethe derivative g(t) is thesquare wave function of Example 6, it follows from the formula in (7b) and Theorem 2 of Section 10.2 that the transform of this triangular wavefunction is

(8)G(s)=F(s)s=1s2tanhas2.

FIGURE 10.5.9.

The graph of the triangular wave function of Example 6.

Example 7

Square wave forcing Consider a mass–spring–dashpot system with m=1, c=4, and k=20 in appropriate units. Suppose that the system is initially at rest at equilibrium (x(0)=x(0)=0) and that the mass is acted on beginning at time t=0 by the external force f(t) whose graph is shown in Fig. 10.5.10: the square wave with amplitude 20 and period 2π. Find the position function f(t).

Solution

The initial value problem is

x+4x+20x=f(t);x(0)=x(0)=0.

The transformed equation is

(9)s2X(s)+4sX(s)+20X(s)=F(s).

FIGURE 10.5.10.

The graph of the external force function of Example 7.

From Example 5 with a=π we see that the transform of f(t) is

F(s)=20s1eπs1+eπs=20s(1eπs)(1eπs+e2πse3πs+)=20s(12eπs+2e2πs2e3πs+),

so that

(10)F(s)=20s+40sn=1(1)nenπs.

Substitution of Eq. (10) in Eq. (9) yields

(11)X(s)=F(s)s2+4s+20=20s[(s+2)2+16]+2n=1(1)n20enπss[(s+2)2+16.

From the transform in Eq. (8) of Section 10.3, we get

L1{20(s+2)2+16}=5e2tsin 4t,

so by Theorem 2 of Section 10.2 we have

g(t)=L1{20s[(s+2)2+16]}=0t5e2τsin 4τdτ.

Using a tabulated formula for eat sin bt dt, we get

(12)g(t)=1e2t(cos 4t+12sin 4t)=1h(t),

where

(13)h(t)=e2t(cos 4t+12sin 4t).

Now we apply Theorem 1 to find the inverse transform of the right-hand term in Eq. (11). The result is

(14)x(t)=g(t)+2n=1(1)nu(tnπ)g(tnπ),

and we note that for any fixed value of t the sum in Eq. (14) is finite. Moreover,

g(tnπ)=1e2(tnπ)[cos 4(tnπ)+12sin 4(tnπ)]=1e2nπe2t(cos 4t+12sin 4t).

Therefore,

(15)g(tnπ)=1e2nπh(t).

Hence if 0<t<π, then

x(t)=1h(t).

If π<t<2π, then

x(t)=[1h(t)]2[1e2πh(t)]=1+h(t)2h(t)[1e2π].

If 2π<t<3π, then

x(t)=[1h(t)]2[1e2πh(t)]+2[1e4πh(t)]=1+h(t)2h(t)[1e2π+e4π].

The general expression for nπ<t<(n+1)π is

(16)x(t)=h(t)+(1)n2h(t)[1e2π++(1)ne2nπ]=h(t)+(1)n2h(t)1+(1)ne2(n+1)π1+e2π,

which we obtained with the aid of the familiar formula for the sum of a finite geometric progression. A rearrangement of Eq. (16) finally gives, with the aid of Eq. (13),

(17)x(t)=e2π1e2π+1e2t(cos 4t+12sin 4t)+(1)n2(1)ne2πe2π+1e2(tnπ)(cos 4t+12sin 4t)

for nπ<t<(n+1)π. The first term in Eq. (17) is the transient solution

(18)xtr(t)(0.9963)e2t(cos 4t+12sin 4t)(1.1139)e2tcos(4t0.4636).

The last two terms in Eq. (17) give the steady periodic solution xsp. To investigate it, we write τ=tnπ for t in the interval nπ<t<(n+1)π. Then

(19)xsp(t)=(1)n[12e2πe2π+1e2τ(cos 4τ+12sin 4τ)](1)n[1(2.2319)e2τcos(4τ0.4636)].

Figure 10.5.11 shows the graph of xsp(t). Its most interesting feature is the appearance of periodically damped oscillations with a frequency four times that of the imposed force f(t).

FIGURE 10.5.11.

The graph of the steady periodic solution for Example 7; note the “periodically damped” oscillations with frequency four times that of the imposed force.

10.5 Problems

Find the inverse Laplace transform f(t) of each function given in Problems 1 through 10. Then sketch the graph of f.

  1. F(s)=e3ss2

  2. F(s)=ese3ss2

  3. F(s)=ess+2

  4. F(s)=ese22ss1

  5. F(s)=eπss2+1

  6. F(s)=sess2+π2

  7. F(s)=1e2πss2+1

  8. F(s)=s(1e2s)s2+π2

  9. F(s)=s(1+e3s)s2+π2

  10. F(s)=2s(eπse2πs)s2+4

Find the Laplace transforms of the functions given in Problems 11 through 22.

  1. f(t)=2 if 0t<3; f(t)=0 if t3

  2. f(t)=1 if 1t4; f(t)=0 if t<1 or if t>4

  3. f(t)=sin t if 0t2π; f(t)=0 if t>2π

  4. f(t)=cos πt if 0t2; f(t)=0 if t>2

  5. f(t)=sin t if 0t3π; f(t)=0 if t>3π

  6. f(t)=sin 2t if πt2π; f(t)=0 if t<π or if t>2π

  7. f(t)=sin πt if 2t3; f(t)=0 if t<2 or if t>3

  8. f(t)=cos12πt if 3t5; f(t)=0 if t<3 or if t>5

  9. f(t)=0 if t<1; f(t)=t if t1

  10. f(t)=t if t1; f(t)=1 if t>1

  11. f(t)=t if t1; f(t)=2t if 1t2; f(t)=0 if t>2

  12. f(t)=t3 if 1t2; f(t)=0 if t<1 or if t>2

  13. Apply Theorem 2 with p=1 to verify that L{1}=1/s.

  14. Apply Theorem 2 to verify that L{cos kt}=s/(s2+k2).

  15. Apply Theorem 2 to show that the Laplace transform of the square wave function of Fig. 10.5.12 is

    L{f(t)}=1s(1+eas).

    FIGURE 10.5.12.

    The graph of the square wave function of Problem 25.

  16. Apply Theorem 2 to show that the Laplace transform of the sawtooth function f(t) of Fig. 10.5.13 is

    F(s)=1as2eass(1eas).

    FIGURE 10.5.13.

    The graph of the sawtooth function of Problem 26.

  17. Let g(t) be the staircase function of Fig. 10.5.14. Show that g(t)=(t/a)f(t), where f is the sawtooth function of Fig. 10.5.14, and hence deduce that

    L{g(t)}=eass(1eas).

    FIGURE 10.5.14.

    The graph of the staircase function of Problem 27.

  18. Suppose that f(t) is a periodic function of period 2a with f(t)=t if 0t<a and f(t)=0 if at<2a. Find L{f(t)}.

  19. Suppose that f(t) is the half-wave rectification of sin kt, shown in Fig. 10.5.15. Show that

    L{f(t)}=k(s2+k2)(1eπs/k).

    FIGURE 10.5.15.

    The half-wave rectification of sin kt.

  20. Let g(t)=u(tπ/k)f(tπ/k), where f(t) is the function of Problem 29 and k>0. Note that h(t)=f(t)+g(t) is the full-wave rectification of sin kt shown in Fig. 10.5.16. Hence deduce from Problem 29 that

    L{h(t)}=ks2+k2cothπs2k.

    FIGURE 10.5.16.

    The full-wave rectification of sin kt.

Discontinuous Forcing

In Problems 31 through 35, the values of mass m, spring constant k, dashpot resistance c, and force f(t) are given for a mass–spring–dashpot system with external forcing function. Solve the initial value problem

mx+cx+kx=f(t);x(0)=x(0)=0

and construct the graph of the position function x(t).

  1. m=1, k=4, c=0; f(t)=1 if 0t<π, f(t)=0 if tπ

  2. m=1, k=4, c=5; f(t)=1 if 0t<2, f(t)=0 if t2

  3. m=1, k=9, c=0; f(t)=sin t if 0t2π, f(t)=0 if t>2π

  4. m=1, k=1, c=0; f(t)=t if 0t<1, f(t)=0 if t1

  5. m=1, k=4, c=4; f(t)=t if 0t2, f(t)=0 if t>2

Transient and Steady Periodic Motions

In Problems 36 and 37, a mass–spring–dashpot system with external force f(t) is described. Under the assumption that x(0)=x(0)=0, use the method of Example 7 to find the transient and steady periodic motions of the mass. Then construct the graph of the position function x(t). If you would like to check your graph using a numerical DE solver, it may be useful to note that the function

f(t)=A[2u((tπ)(t2π)(t3π).(t4π)(t5π)(t6π))1]

has the value +A if 0<t<π, the value A if π<t<2π, and so forth, and hence agrees on the interval [0,6π] with the square wave function that has amplitude A and period 2π. (See also the definition of a square wave function interms of sawtooth and triangular wave functions in the application material for this section.)

  1. m=1, k=4, c=0; f(t) is a square wave function with amplitude 4 and period 2π.

  2. m=1, k=10, c=2; f(t) is a square wave function with amplitude 10 andperiod 2π.

  3. Suppose the function x(t) satisfies the initial value problem

    mx+cx+kx=F(t), x(a)=b0, x(a)=b1

    for ta and x(t)=0 for t<a. Then show that X(s)=L{x(t)} satisfies the equation

    m(s2(easX)sb0b1)+c(s(easX)b0)+k(easX)=L{F(t+a)}.
  4. This is an alternate approach to Example 7 that was suggested by Keng C. Wu of Lockheed Martin (Maritime Systems Sensors). Let x(t) denote the steady periodic solution of the given differential equation x+4x+20x=F(t). Suppose we write v(t) for the restriction of x(t) to the first half [0,π] of the fundamental interval [0,2π], and w(t) for its restriction to the second half-interval [π,2π]. We may regard v(t) as the solution of the initial value problem

    v+4v+20v=+20,v(0)=b0, v(0)=b1

    and w(t) as the solution of the initial value problem

    w+4w+20w=20,w(π)=c0, w(π)=c1,

    where the initial values b0, b1 and c0, c1 are to be determined so that x(t) and x(t) are continuous.

    1. Transform the first of these initial value problems to show that V(s)=A(s)b0+B(s)b1+C(s), where

      A(s)=s+4s2+4s+20,B(s)=1s2+4s+20,C(s)=20s(s2+4s+20).
    2. Apply the result of Problem 38 to the second initial value problem above to show that W(s)=eπs(A(s)c0+B(s)c1C(s)), where the coefficient functions A(s), B(s), C(s) are as defined in part (a).

    3. After finding the inverse transforms

      a(t)=e2t(cos 4t+12sin 4t),b(t)=14e2tsin 4t,c(t)=1e2t(cos 4t+12sin 4t),

      solve the four continuity equations v(π)=c0, v(π)=c1, w(2π)=b0, w(2π)=b1 to find the values of the previously undetermined initial values. Conclude that

      v(t)10.9981e2t(2 cos 4t+sin 4t),w(t)[1+0.9981e2(tπ)(2 cos 4t+sin 4t)]u(tπ).

      Finally, use these expressions to verify that the graph of the steady periodic solution x(t) looks as indicated in Fig. 10.5.11.

10.5 Application Engineering Functions

Periodic piecewise linear functions occur so frequently as input functions in engineering applications that they are sometimes called engineering functions. Computations with such functions are readily handled by computer algebra systems. In Mathematica, for instance, the SawToothWave, TriangleWave, and Square-Wave functions can be used to create the corresponding inputs with specified range, period, etc. Alternatively, we can define our own engineering functions using elementary functions available in any computer algebra system:

sawtooth[t_] := t - 2 Floor[t/2] - 1
triangularwave[t_] := 2 Abs[sawtooth[t - 1/2]] - 1
squarewave[t_] := Sign[ triangularwave[t]]

Plot each of the functions to verify that it has period 2 and that its name is aptly chosen. For instance, the result of

Plot[squarewave[t], {t, 0, 6}]

should look like Fig. 10.5.8. If f(t) is one of these engineering functions and p>0, then the function f(2t/p) will have period p. To illustrate this, try

Plot[triangularwave[ 2 t/p ], {t, 0, 3 p}]

with various values of p. Now let’s consider the mass–spring–dashpot equation

diffEq = m x″ [t] + c x′ [t] + k x[t] == input

with selected parameter values and an input forcing function with period p and amplitude F0.

m = 4; c = 8; k = 5; p = 1; F0 = 4;
input = F0 squarewave[2 t/p];

You can plot this input function to verify that it has period 1:

Plot[input, {t, 0, 2}]

Finally, let’s suppose that the mass is initially at rest in its equilibrium position and solve numerically the resulting initial value problem.

response = NDSolve[ {diffEq, x[0] == 0, x′[0] == 0},
x, {t, 0, 10}]
Plot[ x[t] /. response, {t, 0, 10}]

In the resulting Fig. 10.5.17 we see that after an initial transient dies out, the response function x(t) settles down (as expected?) to a periodic oscillation with the same period as the input.

Investigate this initial value problem with several mass–spring–dashpot para–meters–for instance, selected digits of your student ID number—and with input engineering functions having various amplitudes and periods.

FIGURE 10.5.17.

Response x(t) to period 1 square wave input.

11 Power Series Methods

11.1 Introduction and Review of Power Series

In Section 5.3 we saw that solving a homogeneous linear differential equation with constant coefficients can be reduced to the algebraic problem of finding the roots of its characteristic equation. There is no similar procedure for solving linear differential equations with variable coefficients, at least not routinely and in finitely many steps. With the exception of special types, such as the occasional equation that can be solved by inspection, linear equations with variable coefficients generally require the power series techniques of this chapter.

These techniques suffice for many of the nonelementary differential equations that appear most frequently in applications. Perhaps the most important (because of its applications in such areas as acoustics, heat flow, and electromagnetic radiation) is Bessel’s equation of order n:

x2y+xy+(x2n2)y=0.

Legendre’s equation of order n is important in many applications. It has the form

(1x2)y2xy+n(n+1)y=0.

In this section we introduce the power series method in its simplest form and, along the way, state (without proof) several theorems that constitute a review of the basic facts about power series. Recall first that a power series in (powers of) xa is an infinite series of the form

(1)n=0cn(xa)n=c0+c1(xa)+c2(xa)2++cn(xa)n+.

If a=0, this is a power series in x:

(2)n=0cnxn=c0+c1x+c2x2++cnxn+.

We will confine our review mainly to power series in x, but every general property of power series in x can be converted to a general property of power series in xa by replacement of x with xa.

The power series in (2) converges on the interval I provided that the limit

(3)n=0cnxn=limNn=0Ncnxn

exists for all x in I. In this case the sum

(4)f(x)=n=0cnxn

is defined on I, and we call the series cnxn a power series representation of the function f on I. The following power series representations of elementary functions should be familiar to you from introductory calculus:

(5)ex=n=0xnn!=1+x+x22!+x33!+;
(6)cosx=n=0(1)nx2n(2n)!=1x22!+x44!;
(7)sinx=n=0(1)nx2n+1(2n+1)!=xx33!+x55!;
(8)coshx=n=0x2n(2n)!=1+x22!+x44!+;
(9)sinhx=n=0x2n+1(2n+1)!=x+x33!+x55!+;
(10)ln(1+x)=n=0(1)n+1xnn=xx22+x33;
(11)11x=n=0xn=1+x+x2+x3+;

and

(12)(1+x)α=1+αx+α(α1)x22!+α(α1)(α2)x33!+.

In compact summation notation, we observe the usual conventions that 0!=1 and that x0=1 for all x, including x=0. The series in (5) through (9) converge to the indicated functions for all x. In contrast, the series in (10) and (11) converge if |x|<1 but diverge if |x|>1. (What if |x|=1?) The series in (11) is the geometric series. The series in (12), with α an arbitrary real number, is the binomial series. If α is a nonnegative integer n, then the series in (12) terminates and the binomial series reduces to a polynomial of degree n which converges for all x. Otherwise, the series is actually infinite and it converges if |x|<1 and diverges if |x|>1; its behavior for |x|=1 depends on the value of α.

Remark

Power series such as those listed in (5) through (12) are often derived as Taylor series. The Taylor series with center x=a of the function f is the power series

(13)n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2+

in powers of xa, under the hypothesis that f is infinitely differentiable at x=a (so that the coefficients in Eq. (13) are all defined). If a=0, then the series in (13) is the Maclaurin series

(13′)n=0f(n)(0)n!xn=f(0)+f(0)x+f(0)2!x2+f(3)(0)3!x3+.

For example, suppose that f(x)=ex. Then f(n)(x)=ex, and hence f(n)(0)=1 for all n0. In this case Eq. (13′) reduces to the exponential series in (5).

Power Series Operations

If the Taylor series of the function f converges to f(x) for all x in some open interval containing a, then we say that the function f is analytic at x=a. For example,

  • every polynomial function is analytic everywhere;

  • every rational function is analytic wherever its denominator is nonzero;

  • more generally, if the two functions f and g are both analytic at x=a, then so are their sum f+g and their product f·g, as is their quotient f/g if g(a)0.

For instance, the function h(x)=tanx=(sin x)/(cos x) is analytic at x=0 because cos 0=10 and the sine and cosine functions are analytic (by virtue of their convergent power series representations in Eqs. (6) and (7)). It is rather awkward to compute the Taylor series of the tangent function using Eq. (13) because of the way in which its successive derivatives grow in complexity (try it!). Fortunately, power series may be manipulated algebraically in much the same way as polynomials. For example, if

(14)f(x)=n=0anxnandg(x)=n=0bnxn,

then

(15)f(x)+g(x)=n=0(an+bn)xn

and

(16)f(x)g(x)=n=0cnxn=a0b0+(a0b1+a1b0)x+(a0b2+a1b1+a2b0)x2+,

where cn=a0bn+a1bn1++anb0. The series in (15) is the result of termwise addition, and the series in (16) is the result of formal multiplication—multiplying each term of the first series by each term of the second and then collecting coefficients of like powers of x. (Thus the processes strongly resemble addition and multiplication of ordinary polynomials.) The series in (15) and (16) converge to f(x)+g(x) and f(x)g(x), respectively, on any open interval on which both the series in (14) converge. For example,

sinxcosx=(x16x3+1120x5)(112x2+124x4)=x+(1612)x3+(124+112+1120)x5+=x46x3+16120x5=12[(2x)(2x)33!+(2x)55!]=12sin2x

for all x.

Similarly, the quotient of two power series can be computed by long division, as illustrated by the computation shown in Fig. 11.1.1. This division of the Taylor series for cos x into that for sin x yields the first few terms of the series

(17)tanx=x+13x3+215x5+17315x7+.

Division of power series is more treacherous than multiplication; the series thus obtained for f/g may fail to converge at some points where the series for f and g both converge. For example, the sine and cosine series converge for all x, but the tangent series in (17) converges only if |x|<π/2.

x+x33+2x515+17x7315+1x22+x424x6720+xx36+x5120x75040+xx32+x524x7720+_x33x530+x7840+x33x56+x772_2x5154x7315+2x515x715+_17x7315+

FIGURE 11.1.1.

Obtaining the series for tan x by division of series.

The Power Series Method

The power series method for solving a differential equation consists of substituting the power series

(18)y=n=0cnxn

in the differential equation and then attempting to determine what the coefficients c0, c1, c2,  must be in order that the power series will satisfy the differential equation. This is reminiscent of the method of undetermined coefficients, but now we have infinitely many coefficients somehow to determine. This method is not always successful, but when it is we obtain an infinite series representation of a solution, in contrast to the “closed form” solutions that our previous methods have yielded.

Before we can substitute the power series in (18) in a differential equation, we must first know what to substitute for the derivatives y, y, . The following theorem (stated without proof) tells us that the derivative y of y=cnxn is obtained by the simple procedure of writing the sum of the derivatives of the individual terms in the series for y.

For example, differentiation of the geometric series

(11)11x=n=0xn=1+x+x2+x3+

gives

1(1x)2=n=1nxn1=1+2x+3x2+4x3+.

The process of determining the coefficients in the series y=cnxn so that it will satisfy a given differential equation depends also on Theorem 2. This theorem—also stated without proof—tells us that if two power series represent the same function, then they are the same series. In particular, the Taylor series in (13) is the only power series (in powers of xa) that represents the function f.

In particular, if anxn=0 for all x in some open interval, it follows from Theorem 2 that an=0 for all n0.

Example 1

Solve the equation y+2y=0.

Solution

We substitute the series

y=n=0cnxnandy=n=1ncnxn1

and obtain

(21)n=1ncnxn1+2n=0cnxn=0.

To compare coefficients here, we need the general term in each sum to be the term containing xn. To accomplish this, we shift the index of summation in the first sum. To see how to do this, note that

n=1ncnxn1=c1+2c2x+3c3x2+=n=0(n+1)cn+1xn.

Thus we can replace n with n+1 if, at the same time, we start counting one step lower; that is, at n=0 rather than at n=1. This is a shift of +1 in the index of summation. The result of making this shift in Eq. (21) is the identity

n=0(n+1)cn+1xn+2n=0cnxn=0;

that is,

n=0[(n+1)cn+1+2cn]xn=0.

If this equation holds on some interval, then it follows from the identity principle that (n+1)cn+1+2cn=0 for all n0; consequently,

(22)cn+1=2cnn+1

for all n0. Equation (22) is a recurrence relation from which we can successively compute c1, c2, c3,  in terms of c0; the latter will turn out to be the arbitrary constant that we expect to find in a general solution of a first-order differential equation.

With n=0, Eq. (22) gives

c1=2c01.

With n=1, Eq. (22) gives

c2=2c12=+22c012=22c02!.

With n=2, Eq. (22) gives

c3=2c23=23c0123=23c03!.

By now it should be clear that after n such steps, we will have

cn=(1)n2nc0n!,n1.

(This is easy to prove by induction on n.) Consequently, our solution takes the form

y(x)=n=0cnxn=n=0(1)n2nc0n!xn=c0n=0(2x)nn!=c0e2x.

In the final step we have used the familiar exponential series in Eq. (5) to identify our power series solution as the same solution y(x)=c0e2x we could have obtained by the method of separation of variables.

Shift of Index of Summation

In the solution of Example 1 we wrote

(23)n=1ncnxn1=n=0(n+1)cn+1xn

by shifting the index of summation by +1 in the series on the left. That is, we simultaneously increased the index of summation by 1 (replacing n with n+1, nn+1) and decreased the starting point by 1, from n=1 to n=0, thereby obtaining the series on the right. This procedure is valid because each infinite series in (23) is simply a compact notation for the single series

(24)c1+2c2x+3c3x2+4c4x3+.

More generally, we can shift the index of summation by k in an infinite series by simultaneously increasing the summation index by k (nn+k) and decreasing the starting point by k. For instance, a shift by +2 (nn+2) yields

n=3anxn1=n=1an+2xn+1.

If k is negative, we interpret a “decrease by k” as an increase by k=|k|. Thus a shift by 2 (nn2) in the index of summation yields

n=1ncnxn1=n=3(n2)cn2xn3;

we have decreased the index of summation by 2 but increased the starting point by 2, from n=1 to n=3. You should check that the summation on the right is merely another representation of the series in (24).

We know that the power series obtained in Example 1 converges for all x because it is an exponential series. More commonly, a power series solution is not recognizable in terms of the familiar elementary functions. When we get an unfamiliar power series solution, we need a way of finding where it converges. After all, y=cnxn is merely an assumed form of the solution. The procedure illustrated in Example 1 for finding the coefficients {cn} is merely a formal process and may or may not be valid. Its validity—in applying Theorem 1 to compute y and applying Theorem 2 to obtain a recurrence relation for the coefficients—depends on the convergence of the initially unknown series y=cnxn. Hence this formal process is justified only if in the end we can show that the power series we obtain converges on some open interval. If so, it then represents a solution of the differential equation on that interval. The following theorem (which we state without proof) may be used for this purpose.

The number ρ in (25) is called the radius of convergence of the power series cnxn. For instance, for the power series obtained in Example 1, we have

ρ=limn|(1)n2nc0/n!(1)n+12n+1c0/(n+1)!|=limnn+12=,

and consequently the series we obtained in Example 1 converges for all x. Even if the limit in (25) fails to exist, there always will be a number ρ such that exactly one of the three alternatives in Theorem 3 holds. This number may be difficult to find, but for the power series we will consider in this chapter, Eq. (25) will be quite sufficient for computing the radius of convergence.

Example 2

Solve the equation (x3)y+2y=0.

Solution

As before, we substitute

y=n=0cnxnandy=n=1ncnxn1

to obtain

(x3)n=1ncnxn1+2n=0cnxn=0

so that

n=1ncnxn3n=1ncnxn1+2n=0cnxn=0.

In the first sum we can replace n=1 with n=0 with no effect on the sum. In the second sum we shift the index of summation by +1. This yields

n=0ncnxn3n=0(n+1)cn+1xn+2n=0cnxn=0;

that is,

n=0[ncn3(n+1)cn+1+2cn]xn=0.

The identity principle then gives

ncn3(n+1)cn+1+2cn=0,

from which we obtain the recurrence relation

cn+1=n+23(n+1)cnforn0.

We apply this formula with n=0, n=1, and n=2, in turn, and find that

c1=23c0,c2=332c1=332c0,andc3=433c2=433c0.

This is almost enough to make the pattern evident; it is not difficult to show by induction on n that

cn=n+13nc0ifn1.

Hence our proposed power series solution is

(26)y(x)=c0n=0n+13nxn.

Its radius of convergence is

ρ=limn|cncn+1|=limn3n+3n+2=3.

Thus the series in (26) converges if 3<x<3 and diverges if |x|>3. In this particular example we can explain why. An elementary solution (obtained by separation of variables) of our differential equation is y=1/(3x)2. If we differentiate termwise the geometric series

13x=131x3=13n=0xn3n,

we get a constant multiple of the series in (26). Thus this series (with the arbitrary constant c0 appropriately chosen) represents the solution

y(x)=1(3x)2

on the interval 3<x<3, and the singularity at x=3 is the reason why the radius of convergence of the power series solution turned out to be ρ=3.

Example 3

Solve the equation x2y=yx1.

Solution

We make the usual substitutions y=cnxn and y=ncnxn1, which yield

x2n=1ncnxn1=1x+n=0cnxn

so that

n=1ncnxn+1=1x+n=0cnxn.

Because of the presence of the two terms 1 and x on the right-hand side, we need to split off the first two terms, c0+c1x, of the series on the right for comparison. If we also shift the index of summation on the left by 1 (replace n=1 with n=2 and n with n1), we get

n=2(n1)cn1xn=1x+c0+c1x+n=2cnxn.

Because the left-hand side contains neither a constant term nor a term containing x to the first power, the identity principle now yields c0=1, c1=1, and cn=(n1)cn1 for n2. It follows that

c2=1c1=1!,c3=2c2=2!,c4=3c3=3!,

and, in general, that

cn=(n1)!forn2.

Thus we obtain the power series

y(x)=1+x+n=2(n1)!xn.

But the radius of convergence of this series is

ρ=limn(n1)!n!=limn1n=0,

so the series converges only for x=0. What does this mean? Simply that the given differential equation does not have a (convergent) power series solution of the assumed form y=cnxn. This example serves as a warning that the simple act of writing y=cnxn involves an assumption that may be false.

Example 4

Solve the equation y+y=0.

Solution

If we assume a solution of the form

y=n=0cnxn,

we find that

y=n=1ncnxn1andy=n=2n(n1)cnxn2.

Substitution for y and y in the differential equation then yields

n=2n(n1)cnxn2+n=0cnxn=0.

We shift the index of summation in the first sum by +2 (replace n=2 with n=0 and n with n+2). This gives

n=0(n+2)(n1)cn+2xn+n=0cnxn=0.

The identity (n+2)(n+1)cn+2+cn=0 now follows from the identity principle, and thus we obtain the recurrence relation

(27)cn+2=cn(n+1)(n+2)

for n0. It is evident that this formula will determine the coefficients cn with even subscript in terms of c0 and those of odd subscript in terms of c1; c0 and c1 are not predetermined and thus will be the two arbitrary constants we expect to find in a general solution of a second-order equation.

When we apply the recurrence relation in (27) with n=0, 2, and 4 in turn, we get

c2=c02!,c4=c04!,andc6=c06!.

Taking n=1, 3, and 5 in turn, we find that

c3=c13!,c5=c15!,andc7=c17!.

Again, the pattern is clear; we leave it for you to show (by induction) that for k1,

c2k=(1)kc0(2k)!andc2k+1=(1)kc1(2k+1)!.

Thus we get the power series solution

y(x)=c0(1x22!+x44!x66!+)+c1(xx33!+x55!x77!+);

that is, y(x)=c0cos x+c1sin x. Note that we have no problem with the radius of convergence here; the Taylor series for the sine and cosine functions converge for all x.

The solution of Example 4 can bear further comment. Suppose that we had never heard of the sine and cosine functions, let alone their Taylor series. We would then have discovered the two power series solutions

(28)C(x)=n=0(1)nx2n(2n)!=1x22!+x44!

and

(29)S(x)=n=0(1)nx2n+1(2n+1)!=xx33!+x55!

of the differential equation y+y=0. Both of these power series converge for all x. For instance, the ratio test in Theorem 3 implies convergence for all z of the series (1)nzn/(2n)! obtained from (28) by writing z=x2. Hence it follows that (28) itself converges for all x, as does (by a similar ploy) the series in (29).

It is clear that C(0)=1 and S(0)=0, and termwise differentiation of the two series in (28) and (29) yields

(30)C(x)=S(x)andS(x)=C(x).

Consequently, C(0)=0 and S(0)=1. Thus with the aid of the power series method (all the while knowing nothing about the sine and cosine functions), we have discovered that y=C(x) is the unique solution of

y+y=0

that satisfies the initial conditions y(0)=1 and y(0)=0, and that y=S(x) is the unique solution that satisfies the initial conditions y(0)=0 and y(0)=1. It follows that C(x) and S(x) are linearly independent, and—recognizing the importance of the differential equation y+y=0—we can agree to call C the cosine function and S the sine function. Indeed, all the usual properties of these two functions can be established, using only their initial values (at x=0) and the derivatives in (30); there is no need to refer to triangles or even to angles. (Can you use the series in (28) and (29) to show that [C(x)]2+[S(x)]2=1 for all x?) This demonstrates that

The cosine and sine functions are fully determined by the differential equation y+y=0 of which they are the two natural linearly independent solutions.

Figures 11.1.2 and 11.1.3 show how the geometric character of the graphs of cos x and sin x is revealed by the graphs of the Taylor polynomial approximations that we get by truncating the infinite series in (28)and (29).

This is by no means an uncommon situation. Many important special functions of mathematics occur in the first instance as power series solutions of differential equations and thus are in practice defined by means of these power series. In the remaining sections of this chapter we will see numerous examples of such functions.

FIGURE 11.1.2.

Taylor polynomial approximations to cos x.

FIGURE 11.1.3.

Taylor polynomial approximations to sin x.

11.1 Problems

In Problems 1 through 10, find a power series solution of the given differential equation. Determine the radius of convergence of the resulting series, and use the series in Eqs. (5) through (12) to identify the series solution in terms of familiar elementary functions. (Of course, no one can prevent you from checking your work by also solving the equations by the methods of earlier chapters!)

  1. y=y

  2. y=4y

  3. 2y+3y=0

  4. y+2xy=0

  5. y=x2y

  6. (x2)y+y=0

  7. (2x1)y+2y=0

  8. 2(x+1)y=y

  9. (x1)y+2y=0

  10. 2(x1)y=3y

In Problems 11 through 14, use the method of Example 4 to find two linearly independent power series solutions of the given differential equation. Determine the radius of convergence of each series, and identify the general solution in terms of familiar elementary functions.

  1. y=y

  2. y=4y

  3. y+9y=0

  4. y+y=x

Show (as in Example 3) that the power series method fails to yield a power series solution of the form y=cnxn for the differential equations in Problems 15 through 18.

  1. xy+y=0

  2. 2xy=y

  3. x2y+y=0

  4. x3y=2y

In Problems 19 through 22, first derive a recurrence relation giving cn for n2 in terms of c0 or c1 (or both). Then apply the given initial conditions to find the values of c0 and c1. Next determine cn (in terms of n, as in the text) and, finally, identify the particular solution in terms of familiar elementary functions.

  1. y+4y=0; y(0)=0, y(0)=3

  2. y4y=0; y(0)=2, y(0)=0

  3. y2y+y=0; y(0)=0, y(0)=1

  4. y+y2y=0; y(0)=1, y(0)=2

  5. Show that the equation

    x2y+x2y+y=0

    has no power series solution of the form y=cnxn.

  6. Establish the binomial series in (12) by means of the following steps. (a) Show that y=(1+x)α satisfies the initial value problem (1+x)y=αy, y(0)=1.(b) Show that the power series method gives the binomial series in (12) as the solution of the initial value problem in part (a), and that this series converges if |x|<1. (c) Explain why the validity of the binomial series given in (12) follows from parts (a) and (b).

  7. For the initial value problem

    y=y+y,y(0)=0,y(1)=1

    derive the power series solution

    y(x)=n=1Fnn!xn

    where {Fn}n=0 is the sequence 0, 1, 1, 2, 3, 5, 8, 13, of Fibonacci numbers defined by F0=0, F1=1, Fn=Fn2+Fn1 for n>1.

    1. Show that the solution of the initial value problem

      y=1+y2,y(0)=0

      is y(x)=tanx.

    2. Because y(x)=tanx is an odd function with y(0)=1, its Taylor series is of the form

      y=x+c3x3+c5x5+c7x7+.

      Substitute this series in y=1+y2 and equate like powers of x to derive the following relations:

      3c3=1,5c5=2c3,7c7=2c5+(c3)2,9c9=2c7+2c3c5,11c11=2c9+2c3c7+(c5)2.
    3. Conclude that

      tanx=x+13x3+215x5+17315x7+622835x9+1382155925x11+.
    4. Would you prefer to use the Maclaurin series formula in (13) to derive the tangent series in part (c)? Think about it!

  8. This section introduces the use of infinite series to solve differential equations. Conversely, differential equations can sometimes be used to sum infinite series. For example, consider the infinite series

    1+11!12!+13!+14!15!+;

    note the + + + + pattern of signs superimposed on the terms of the series for the number e. We could evaluate this series if we could obtain a formula for the function

    f(x)=1+x12!x2+13!x3+14!x4+15!x5+,

    because the sum of the numerical series in question is simply f(1).

    1. It’s possible to show that the power series given here converges for all x and that termwise differentiation is valid. Given these facts, show that f(x) satisfies the initial value problem

      y(3)=y;y(0)=y(0)=1,y(0)=1.
    2. Solve this initial value problem to show that

      f(x)=13ex+23ex/2(cos32x+3sin32x).

      For a suggestion, see Problem 48 of Section 5.3.

    3. Evaluate f(1) to find the sum of the numerical series given here.

11.2 Power Series Solutions

The power series method introduced in Section 11.1 can be applied to linear equations of any order (as well as to certain nonlinear equations), but its most important applications are to homogeneous second-order linear differential equations of the form

(1)A(x)y+B(x)y+C(x)y=0,

where the coefficients A, B, and C are analytic functions of x. Indeed, in most applications these coefficient functions are simple polynomials.

We saw in Example 3 of Section 11.1 that the series method does not always yield a series solution. To discover when it does succeed, we rewrite Eq. (1) in the form

(2)y+P(x)y+Q(x)y=0

with leading coefficient 1, and with P=B/A and Q=C/A. Note that P(x) and Q(x) will generally fail to be analytic at points where A(x) vanishes. For instance, consider the equation

(3)xy+y+xy=0.

The coefficient functions in (3) are continuous everywhere. But in the form of (2) it is the equation

(4)y+1xy+y=0

with P(x)=1/x not analytic at x=0.

The point x=a is called an ordinary point of Eq. (2)—and of the equivalent Eq. (1)—provided that the functions P(x) and Q(x) are both analytic at x=a. Otherwise, x=a is a singular point. Thus the only singular point of Eqs. (3) and (4) is x=0. Recall that a quotient of analytic functions is analytic wherever the denominator is nonzero. It follows that, if A(a)0 in Eq. (1) with analytic coefficients, then x=a is an ordinary point. If A(x), B(x), and C(x) are polynomials with no common factors, then x=a is an ordinary point if and only if A(a)0.

Example 1

The point x=0 is an ordinary point of the equation

xy+(sinx)y+x2y=0,

despite the fact that A(x)=x vanishes at x=0. The reason is that

P(x)=sinxx=1x(xx33!+x55!)=1x23!+x45!

is nevertheless analytic at x=0 because the division by x yields a convergent power series.

Example 2

The point x=0 is not an ordinary point of the equation

y+x2y+x1/2y=0.

For while P(x)=x2 is analytic at the origin, Q(x)=x1/2 is not. The reason is that Q(x) is not differentiable at x=0 and hence is not analytic there. (Theorem 1 of Section 11.1 implies that an analytic function must be differentiable.)

Example 3

The point x=0 is an ordinary point of the equation

(1x3)y+(7x2+3x5)y+(5x13x4)y=0

because the coefficient functions A(x), B(x), and C(x) are polynomials with A(0)0.

Theorem 2 of Section 3.1 implies that Eq. (2) has two linearly independent solutions on any open interval where the coefficient functions P(x) and Q(x) are continuous. The basic fact for our present purpose is that near an ordinary point a, these solutions will be power series in powers of xa. A proof of the following theorem can be found in Chapter 3 of Coddington, An Introduction to Ordinary Differential Equations (Dover Publications, 1989).

Example 4

Determine the radius of convergence guaranteed by Theorem 1 of a series solution of

(6)(x2+9)y+xy+x2y=0

in powers of x. Repeat for a series in powers of x4.

Solution

This example illustrates the fact that we must take into account complex singular points as well as real ones. Because

P(x)=xx2+9andQ(x)=x2x2+9,

the only singular points of Eq. (6) are ±3i. The distance (in the complex plane) of each of these from 0 is 3, so a series solution of the form cnxn has radius of convergence at least 3. The distance of each singular point from 4 is 5, so a series solution of the form cn(x4)n has radius of convergence at least 5 (see Fig. 11.2.1).

Example 5

Find the general solution in powers of x of

(7)(x24)y+3xy+y=0.

Then find the particular solution with y(0)=4, y(0)=1.

Solution

The only singular points of Eq. (7) are ±2, so the series we get will have radius of convergence at least 2. (See Problem 35 for the exact radius of convergence.) Substitution of

y=n=0cnxn,y=n=1ncnxn1,andy=n=2n(n1)cnxn2

in Eq. (7) yields

n=2n(n1)cnxn4n=2n(n1)cnxn2+3n=1ncnxn+n=0cnxn=0.

FIGURE 11.2.1.

Radius of convergence as distance to nearest singularity.

We can begin the first and third summations at n=0 as well, because no nonzero terms are thereby introduced. We shift the index of summation in the second sum by +2, replacing n with n+2 and using the initial value n=0. This gives

n=0n(n1)cnxn4n=0(n+2)(n+1)cn+2xn+3n=0ncnxn+n=0cnxn=0.

After collecting coefficients of cn and cn+2, we obtain

n=0[(n2+2n+1)cn4(n+2)(n+1)cn+2]xn=0.

The identity principle yields

(n+1)2cn4(n+2)(n+1)cn+2=0,

which leads to the recurrence relation

(8)cn+2=(n+1)cn4(n+2)

for n0. With n=0, 2, and 4 in turn, we get

c2=c042,c4=3c244=3c04224,andc6=5c446=35c043246.

Continuing in this fashion, we evidently would find that

c2n=135(2n1)4n24(2n)c0.

With the common notation

(2n+1)!!=135(2n+1)=(2n+1)!2nn!

and the observation that 2·4·6(2n)=2n·n!, we finally obtain

(9)c2n=(2n1)!!23nn!c0.

(We also used the fact that 4n·2n=23n.)

With n=1, 3, and 5 in Eq. (8), we get

c3=2c143,c5=4c345=24c14235,andc7=6c547=246c143357.

It is apparent that the pattern is

(10)c2n+1=246(2n)4n135(2n+1)c1=n!2n(2n+1)!!c1.

The formula in (9) gives the coefficients of even subscript in terms of c0; the formula in (10) gives the coefficients of odd subscript in terms of c1. After we separately collect the terms of the series of even and odd degree, we get the general solution

(11)y(x)=c0(1+n=1(2n1)!!23nn!x2n)+c1(x+n=1n!2n(2n+1)!!x2n+1).

Alternatively,

(11′)y(x)=c0(1+18x2+3128x4+51024x6+)+c1(x+16x3+130x5+1140x7+).

Because y(0)=c0 and y(0)=c1, the given initial conditions imply that c0=4 and c1=1. Using these values in Eq. (11′), the first few terms of the particular solution satisfying y(0)=4 and y(0)=1 are

(12)y(x)=4+x+12x2+16x3+332x4+130x5+.

Remark

As in Example 5, substitution of y=cnxn in a linear second-order equation with x=0 an ordinary point typically leads to a recurrence relation that can be used to express each of the successive coefficients c2, c3, c4,  in terms of the first two, c0 and c1. In this event two linearly independent solutions are obtained as follows. Let y0(x) be the solution obtained with c0=1 and c1=0, and let y1(x) be the solution obtained with c0=0 and c1=1. Then

y0(0)=1,y0(0)=0andy1(0)=0,y1(0)=1,

so it is clear that y0 and y1 are linearly independent. In Example 5, y0(x) and y1(x) are defined by the two series that appear on the right-hand side in Eq. (11), which expresses the general solution in the form y=c0y0+c1y1.

Translated Series Solutions

If in Example 5 we had sought a particular solution with given initial values y(a) and y(a), we would have needed the general solution in the form

(13)y(x)=n=0cn(xa)n;

that is, in powers of xa rather than in powers of x. For only with a solution of the form in (13) is it true that the initial conditions

y(a)=c0andy(a)=c1

determine the arbitrary constants c0 and c1 in terms of the initial values of y and y. Consequently, to solve an initial value problem, we need a series expansion of the general solution centered at the point where the initial conditions are specified.

Example 6

Solve the initial value problem

(14)(t22t3)d2ydt2+3(t1)dydt+y=0;y(1)=4,y(1)=1.

Solution

We need a general solution of the form cn(t1)n. But instead of substituting this series in (14) to determine the coefficients, it simplifies the computations if we first make the substitution x=t1, so that we wind up looking for a series of the form cnxn after all. To transform Eq. (14) into one with the new independent variable x, we note that

t22t3=(x+1)22(x+1)3=x24,dydt=dydxdxdt=dydx=y,

and

d2ydt2=[ddx(dydx)]dxdt=ddx(y)=y,

where primes denote differentiation with respect to x. Hence we transform Eq. (14) into

(x24)y+3xy+y=0

with initial conditions y=4 and y=1 at x=0 (corresponding to t=1). This is the initial value problem we solved in Example 5, so the particular solution in (12) is available. We substitute t1 for x in Eq. (12) and thereby obtain the desired particular solution

y(t)=4+(t1)+12(t1)2+16(t1)3+332(t1)4+130(t1)5+.

This series converges if 1<t<3. (Why?) A series such as this can be used to estimate numerical values of the solution. For instance,

y(0.8)=4+(0.2)+12(0.2)2+16(0.2)3+332(0.2)4+130(0.2)5+,

so that y(0.8)3.8188.

The last computation in Example 6 illustrates the fact that series solutions of differential equations are useful not only for establishing general properties of a solution, but also for numerical computations when an expression of the solution in terms of elementary functions is unavailable.

Types of Recurrence Relation

The formula in Eq. (8) is an example of a two-term recurrence relation; it expresses each coefficient in the series in terms of one of the preceding coefficients. A many-term recurrence relation expresses each coefficient in the series in terms of two or more preceding coefficients. In the case of a many-term recurrence relation, it is generally inconvenient or even impossible to find a formula that gives the typical coefficient cn in terms of n. The next example shows what we sometimes can do with a three-term recurrence relation.

Example 7

Find two linearly independent solutions of

(15)yxyx2y=0.

Solution

We make the usual substitution of the power series y=cnxn. This results in the equation

n=2n(n1)cnxn2n=1ncnxnn=0cnxn+2=0.

We can start the second sum at n=0 without changing anything else. To make each term include xn in its general term, we shift the index of summation in the first sum by +2 (replace n with n+2), and we shift it by 2 in the third sum (replace n with n2). These shifts yield

n=0(n+2)(n+1)cn+2xnn=0ncnxnn=2cn2xn=0.

The common range of these three summations is n2, so we must separate the terms corresponding to n=0 and n=1 in the first two sums before collecting coefficients of xn. This gives

2c2+6c3xc1x+n=2[(n+2)(n+1)cn+2ncncn2]xn=0.

The identity principle now implies that 2c2=0, that c3=16c1, and the three-term recurrence relation

(16)cn+2=ncn+cn2(n+2)(n+1)

for n2. In particular,

(17)c4=2c2+c012,c5=3c3+c120,c6=4c4+c230,c7=5c5+c342,c8=6c6+c456.

Thus all values of cn for n4 are given in terms of the arbitrary constants c0 and c1 because c2=0 and c3=16c1.

To get our first solution y1 of Eq. (15), we choose c0=1 and c1=0, so that c2=c3=0. Then the formulas in (17) yield

c4=112,c5=0,c6=190,c7=0,c8=31120;

thus

(18)y1(x)=1+112x4+190x6+31120x8+.

Because c1=c3=0, it is clear from Eq. (16) that this series contains only terms of even degree.

To obtain a second linearly independent solution y2 of Eq. (15), we take c0=0 and c1=1, so that c2=0 and c3=16. Then the formulas in (17) yield

c4=0,c5=340,c6=0,c7=131008,

so that

(19)y2(x)=x+16x3+340x5+131008x7+.

Because c0=c2=0, it is clear from Eq. (16) that this series contains only terms of odd degree. The solutions y1(x) and y2(x) are linearly independent because y1(0)=1 and y1(0)=0, whereas y2(0)=0 and y2(0)=1. A general solution of Eq. (15) is a linear combination of the power series in (18) and (19). Equation (15) has no singular points, so the power series representing y1(x) and y2(x) converge for all x.

The Legendre Equation

The Legendre equation of order α is the second-order linear differential equation

(20)(1x2)y2xy+α(α+1)y=0,

where the real number α satisfies the inequality α>1. This differential equation has extensive applications, ranging from numerical integration formulas (such as Gaussian quadrature) to the problem of determining the steady-state temperature within a solid spherical ball when the temperatures at points of its boundary are known. The only singular points of the Legendre equation are at +1 and 1, so it has two linearly independent solutions that can be expressed as power series in powers of x with radius of convergence at least 1. The substitution y=cmxm in Eq. (20) leads (see Problem 31) to the recurrence relation

(21)cm+2=(αm)(α+m+1)(m+1)(m+2)cm

for m0. We are using m as the index of summation because we have another role for n to play.

In terms of the arbitrary constants c0 and c1, Eq. (21) yields

c2=α(α+1)2!c0,c3=(α1)(α+2)3!c1,c4=α(α2)(α+1)(α+3)4!c0,c5=(α1)(α3)(α+2)(α+4)5!c1.

We can show without much trouble that for m>0,

(22)c2m=(1)mα(α2)(α4)(α2m+2)(α+1)(α+3)(α+2m1)(2m)!c0

and

(23)c2m+1=(1)m(α1)(α3)(α2m+1)(α+2)(α+4)(α+2m)(2m+1)!c1.

Alternatively,

c2m=(1)ma2mc0andc2m+1=(1)ma2m+1c1,

where a2m and a2m+1 denote the fractions in Eqs. (22) and (23), respectively. With this notation, we get two linearly independent power series solutions

(24)y1(x)=c0m=0(1)ma2mx2mandy2(x)=c1m=0(1)ma2m+1x2m+1

of Legendre’s equation of order α.

Now suppose that α=n, a nonnegative integer. If α=n is even, we see from Eq. (22) that a2m=0 when 2m>n. In this case, y1(x) is a polynomial of degree n and y2 is a (nonterminating) infinite series. If α=n is an odd positive integer, we see from Eq. (23) that a2m+1=0 when 2m+1>n. In this case, y2(x) is a polynomial of degree n and y1 is a (nonterminating) infinite series. Thus in either case, one of the two solutions in (24) is a polynomial and the other is a nonterminating series.

With an appropriate choice (made separately for each n) of the arbitrary constants c0 (n even) or c1 (n odd), the nth-degree polynomial solution of Legendre’s equation of order n,

(25)(1x2)y2xy+n(n+1)y=0,

is denoted by Pn(x) and is called the Legendre polynomial of degree n. It is customary (for a reason indicated in Problem 32) to choose the arbitrary constant so that the coefficient of xn in Pn(x) is (2n)!/[2n(n!)2]. It then turns out that

(26)Pn(x)=k=0N(1)k(2n2k)!2nk!(nk)!(n2k)!xn2k,

where N=n/2, the integral part of n/2. The first six Legendre polynomials are

P0(x)1,P1(x)=x,P2(x)=12(3x21),P3(x)=12(5x33x),P4(x)=18(35x430x2+3),P5(x)=18(63x570x3+15x),

and their graphs are shown in Fig. 11.2.2.

FIGURE 11.2.2.

Graphs y=Pn(x) of the Legendre polynomials for n=1, 2, 3, 4, and 5. The graphs are distinguished by the fact that all n zeros of Pn(x) lie in the interval 1<x<1.

11.2 Problems

Find general solutions in powers of x of the differential equations in Problems 1 through 15. State the recurrence relation and the guaranteed radius of convergence in each case.

  1. (x21)y+4xy+2y=0

  2. (x2+2)y+4xy+2y=0

  3. y+xy+y=0

  4. (x2+1)y+6xy+4y=0

  5. (x23)y+2xy=0

  6. (x21)y6xy+12y=0

  7. (x2+3)y7xy+16y=0

  8. (2x2)yxy+16y=0

  9. (x21)y+8xy+12y=0

  10. 3y+xy4y=0

  11. 5y2xy+10y=0

  12. yx2y3xy=0

  13. y+x2y+2xy=0

  14. y+xy=0 (an Airy equation)

  15. y+x2y=0

Use power series to solve the initial value problems in Problems 16 and 17.

  1. (1+x2)y+2xy2y=0; y(0)=0, y(0)=1

  2. y+xy2y=0; y(0)=1, y(0)=0

Solve the initial value problems in Problems 18 through 22. First make a substitution of the form t=xa, then find a solution cntn of the transformed differential equation. State the interval of values of x for which Theorem 1 of this section guarantees convergence.

  1. y+(x1)y+y=0; y(1)=2, y(1)=0

  2. (2xx2)y6(x1)y4y=0; y(1)=0, y(1)=1

  3. (x26x+10)y4(x3)y+6y=0; y(3)=2, y(3)=0

  4. (4x2+16x+17)y=8y; y(2)=1, y(2)=0

  5. (x2+6x)y+(3x+9)y3y=0; y(3)=0, y(3)=2

In Problems 23 through 26, find a three-term recurrence relation for solutions of the form y=cnxn. Then find the first three nonzero terms in each of two linearly independent solutions.

  1. y+(1+x)y=0

  2. (x21)y+2xy+2xy=0

  3. y+x2y+x2y=0

  4. (1+x3)y+x4y=0

  5. Solve the initial value problem

    y+xy+(2x2+1)y=0;y(0)=1,y(0)=1.

    Determine sufficiently many terms to compute y(1/2) accurate to four decimal places.

In Problems 28 through 30, find the first three nonzero terms in each of two linearly independent solutions of the form y=cnxn. Substitute known Taylor series for the analytic functions and retain enough terms to compute the necessary coefficients.

  1. y+exy=0

  2. (cosx)y+y=0

  3. xy+(sinx)y+xy=0

  4. Derive the recurrence relation in (21) for the Legendre equation.

  5. Follow the steps outlined in this problem to establish Rodrigues’s formula

    Pn(x)=1n!2ndndxn(x21)n

    for the nth-degree Legendre polynomial.

    1. Show that v=(x21)n satisfies the differential equation

      (1x2)v+2nxv=0.

      Differentiate each side of this equation to obtain

      (1x2)v+2(n1)xv+2nv=0.
    2. Differentiate each side of the last equation n times in succession to obtain

      (1x2)v(n+2)2xv(n+1)+n(n+1)v(n)=0.

      Thus u=v(n)=Dn(x21)n satisfies Legendre’s equation of order n.

    3. Show that the coefficient of xn in u is (2n)!/n!; then state why this proves Rodrigues’ formula. (Note that the coefficient of xn in Pn(x) is (2n)!/[2n(n!)2].)

  6. The Hermite equation of order α is

    y2xy+2y=0.
    1. Derive the two power series solutions

      y1=1+m=1(1)m2mα(α2)(α2m+2)(2m)!x2m

      and

      y2=x+m=1(1)m2m(α1)(α3)(α2m+1)(2m+1)!x2m+1.

      Show that y1 is a polynomial if α is an even integer, whereas y2 is a polynomial if α is an odd integer.

    2. The Hermite polynomial of degree n is denoted by Hn(x). It is the nth-degree polynomial solution of Hermite’s equation, multiplied by a suitable constant so that the coefficient of xn is 2n. Show that the first six Hermite polynomials are

      H0(x)1,H1(x)=2x,H2(x)=4x22,H3(x)=8x312x,H4(x)=16x448x2+12,H5(x)=32x5160x3+120x.

      A general formula for the Hermite polynomials is

      Hn(x)=(1)nex2dndxn(ex2).

      Verify that this formula does in fact give an nth-degree polynomial. It is interesting to use a computer algebra system to investigate the conjecture that (for each n) the zeros of the Hermite polynomials Hn and Hn+1 are “interlaced”—that is, the n zeros of Hn lie in the n bounded open intervals whose endpoints are successive pairs of zeros of Hn+1.

  7. The discussion following Example 4 in Section 11.1 suggests that the differential equation y+y=0 could be used to introduce and define the familiar sine and cosine functions. In a similar fashion, the Airy equation

    y=xy

    serves to introduce two new special functions that appear in applications ranging from radio waves to molecular vibrations. Derive the first three or four terms of two different power series solutions of the Airy equation. Then verify that your results agree with the formulas

    y1(x)=1+k=114(3k2)(3k)!x3k

    and

    y2(x)=x+k=125(3k1)(3k+1)!x3k+1

    for the solutions that satisfy the initial conditions y1(0)=1, y1(0)=0 and y2(0)=0, y2(0)=1, respectively. The special combinations

    Ai(x)=y1(x)32/3Γ(23)y2(x)31/3Γ(13)

    and

    Bi(x)=y1(x)31/6Γ(23)+y2(x)31/6Γ(13)

    define the standard Airy functions that appear in mathematical tables and computer algebra systems. Their graphs shown in Fig. 11.2.3 exhibit trigonometric-like oscillatory behavior for x<0, whereas Ai(x) decreases exponentially and Bi(x) increases exponentially as x+. It is interesting to use a computer algebra system to investigate how many terms must be retained in the y1- and y2-series above to produce a figure that is visually indistinguishable from Fig. 11.2.3 (which is based on high-precision approximations to the Airy functions).

    FIGURE 11.2.3.

    The Airy function graphs y=Ai(x) and y=Bi(x).

    1. To determine the radius of convergence of the series solution in Example 5, write the series of terms of even degree in Eq. (11) in the form

      y0(x)=1+n=1c2nx2n=1+n=1anzn

      where an=c2n and z=x2. Then apply the recurrence relation in Eq. (8) and Theorem 3 in Section 11.1 to show that the radius of convergence of the series in z is 4. Hence the radius of convergence of the series in x is 2. How does this corroborate Theorem 1 in this section?

    2. Write the series of terms of odd degree in Eq. (11) in the form

      y1(x)=x(1+n=1c2n+1x2n)=x(1+n=1bnzn)

      to show similarly that its radius of convergence (as a power series in x) is also 2.

11.2 Application Automatic Computation of Series Coefficients

Repeated application of a recurrence relation to grind out successive coefficients is—especially in the case of a recurrence relation with three or more terms—a tedious aspect of the infinite series method. Here we illustrate the use of a computer algebra system not only to automate this task, but also to explore interactively the graphical effect of changing the number k of terms we include in a partial-sum approximation to the actual solution given by the full infinite series. In Example 7 we saw that the coefficients in the series solution y=cnxn of the differential equation

(1)yxyx2y=0

are given in terms of the two arbitrary coefficients c0 and c1 by

(2)c2=0,c3=c16,andcn+2=ncn+cn2(n+2)(n+1)forn2.

It would appear to be a routine matter to implement such a recurrence relation, but a twist results from the fact that a typical computer system array is indexed by the subscripts 1, 2, 3, …, rather than by the subscripts 0, 1, 2, … that match the exponents in the successive terms of a power series that begins with a constant term. For this reason we first rewrite our proposed power series solution in the form

(3)y=n=0cnxn=n=1bnxn1

where bn=cn1 for each n1. Then the first two conditions in (1) imply that b3=0 and b4=16b2; also, the recurrence relation (with n replaced with n1) yields the new recurrence relation

(4)bn+2=cn+1=(n1)cn1+cn3(n+1)n=(n1)bn+bn2n(n+1).

Now we are ready to begin. Suppose that we want to calculate the terms through the 10th degree in (2)with the initial conditions b1=b2=1. Then either the Maple commands


k := 11:          # k terms
b := array(1..k):
b[1] := 1:        # arbitrary
b[2] := 1:        # arbitrary
b[3] := 0:
b[4] := b[2]/6:
for n from 3 by 1 to k − 2 do
   b[n+2] := ((n−1) b[n] + b[n−2])/(n (n+1));
   od;

or the Mathematica commands


k = 11;            (∗ k terms ∗)
b = Table[0, {n,1,k}];
b[[1]] = 1;        (∗ arbitrary ∗)
b[[2]] = 1;        (∗ arbitrary ∗)
b[[3]] = 0;
b[[4]] = b[[2]]/6;
For[n=3, n<=k−2,
   b[[n+2]]=((n−1) b[[n]] + b[[n−2]])/(n (n+1)); n=n+1];

quickly yield the coefficients {bn} corresponding to the solution

(5)y(x)=1+x+x36+x412+3x540+x690+13x71008+3x81120+119x951840+41x10113400+.

You might note that the even- and odd-degree terms here agree with those shown in Eqs. (18) and (19), respectively, of Example 7.

The Matlab commands


k = 11;                     % k terms
b = 0∗(1:k);
b(1) = 1;                   % arbitrary
b(2) = 1;                   % arbitrary
b(3) = 0;
b(4) = b(2)/6;
for n = 3:k−2
   b(n+2) = ((n−1) ∗b(n) + ∗b(n−2))/(n∗ (n+1));
   end
format rat, b

give the same results, except that the coefficient b10 of x9 is shown as 73/31801 rather than the correct value 119/51840 shown in Eq. (4). It happens that

73318010.0022955253while119518400.0022955247,

so the two rational fractions agree when rounded to 9 decimal places. The explanation is that (unlike Mathematica and Maple) Matlab works internally with decimal rather than exact arithmetic. But at the end its format rat algorithm converts a correct 14-place approximation for b10 into an incorrect rational fraction that’s “close but no cigar.”

The Matlab commands above form the basis for the interactive display shown in Fig. 11.2.4, which graphs the actual solution (blue curve)of the differential equation (1), with initial conditions b1=b2=1, together with the approximate solution (black curve) consisting of the terms through the fourth degree (k=5) in (5). The pop-up menu allows the user to vary the number of terms k and thus immediately see the graphical effect of changing the number of terms included in the series expansion. With k=10, the actual and approximate solutions are indistinguishable in this viewing window.

FIGURE 11.2.4.

Matlab interactive display. The blue curve represents the actual solution of the initial value problem yxyx2y=0, y(0)=y(0)=1, whereas the black curve shows the partial-sum approximation of the series solution (5) with terms through the fourth degree (k=5).

Finally, you can substitute b1=1, b2=0 and b1=0, b2=1 separately (instead of b1=b2=1) in the commands shown here to derive partial sums of the two linearly independent solutions displayed in Eqs. (18) and (19) of Example 7. This technique can be applied to any of the examples and problems in this section.

11.3 Frobenius Series Solutions

We now investigate the solution of the homogeneous second-order linear equation

(1)A(x)y+B(x)y+C(x)y=0

near a singular point. Recall that if the functions A, B, and C are polynomials having no common factors, then the singular points of Eq. (1) are simply those points where A(x)=0. For instance, x=0 is the only singular point of the Bessel equation of order n,

x2y+xy+(x2n2)y=0,

whereas the Legendre equation of order n,

(1x2)y2xy+n(n+1)y=0,

has the two singular points x=1 and x=1. It turns out that some of the features of the solutions of such equations of the most importance for applications are largely determined by their behavior near their singular points.

We will restrict our attention to the case in which x=0 is a singular point of Eq. (1). A differential equation having x=a as a singular point is easily transformed by the substitution t=xa into one having a corresponding singular point at 0. For example, let us substitute t=x1 into the Legendre equation of order n. Because

y=dydx=dydtdtdx=dydt,y=d2ydx2=[ddt(dydx)]dtdx=d2ydt2,

and 1x2=1(t+1)2=2tt2, we get the equation

t(t+2)d2ydt22(t+1)dydt+n(n+1)y=0.

This new equation has the singular point t=0 corresponding to x=1 in the original equation; it also has the singular point t=2 corresponding to x=1.

Types of Singular Points

A differential equation having a singular point at 0 ordinarily will not have power series solutions of the form y(x)=cnxn, so the straightforward method of Section 11.2 fails in this case. To investigate the form that a solution of such an equation might take, we assume that Eq. (1) has analytic coefficient functions and rewrite it in the standard form

(2)y+P(x)y+Q(x)y=0,

where P=B/A and Q=C/A. Recall that x=0 is an ordinary point (as opposed to a singular point) of Eq. (2) if the functions P(x) and Q(x) are analytic at x=0; that is, if P(x) and Q(x) have convergent power series expansions in powers of x on some open interval containing x=0. Now it can be proved that each of the functions P(x) and Q(x) either is analytic or approaches ± as x0. Consequently, x=0 is a singular point of Eq. (2) provided that either P(x) or Q(x) (or both) approaches ± as x0. For instance, if we rewrite the Bessel equation of order n in the form

y+1xy+(1n2x2)y=0,

we see that P(x)=1/x and Q(x)=1(n/x)2 both approach infinity as x0.

We will see presently that the power series method can be generalized to apply near the singular point x=0 of Eq. (2), provided that P(x) approaches infinity no more rapidly than 1/x, and Q(x) no more rapidly than 1/x2, as x0. This is a way of saying that P(x) and Q(x) have only “weak” singularities at x=0. To state this more precisely, we rewrite Eq. (2) in the form

(3)y+p(x)xy+q(x)x2y=0,

where

(4)p(x)=xP(x)andq(x)=x2Q(x).

In particular, the singular point x=0 is a regular singular point if p(x) and q(x) are both polynomials. For instance, we see that x=0 is a regular singular point of Bessel’s equation of order n by writing that equation in the form

y+1xy+x2n2x2y=0,

noting that p(x)1 and q(x)=x2n2 are both polynomials in x.

By contrast, consider the equation

2x3y+(1+x)y+3xy=0,

which has the singular point x=0. If we write this equation in the form of (3), we get

y+(1+x)/(2x2)xy+32x2y=0.

Because

p(x)=1+x2x2=12x2+12x

as x0 (although q(x)32 is a polynomial), we see that x=0 is an irregular singular point. We will not discuss the solution of differential equations near irregular singular points; this is a considerably more advanced topic than the solution of differential equations near regular singular points.

Example 1

Consider the differential equation

x2(1+x)y+x(4x2)y+(2+3x)y=0.

In the standard form y+Py+Qy=0 it is

y+4x2x(1+x)y+2+3xx2(1+x)y=0.

Because

P(x)=4x2x(1+x)andQ(x)=2+3xx2(1+x)

both approach as x0, we see that x=0 is a singular point. To determine the nature of this singular point, we write the differential equation in the form of Eq. (3):

y+(4x2)/(1+x)xy+(2+3x)/(1+x)x2y=0.

Thus

p(x)=4x21+xandq(x)=2+3x1+x.

Because a quotient of polynomials is analytic wherever the denominator is nonzero, we see that p(x) and q(x) are both analytic at x=0. Hence x=0 is a regular singular point of the given differential equation.

It may happen that when we begin with a differential equation in the general form in Eq. (1) and rewrite it in the form in (3), the functions p(x) and q(x) as given in (4) are indeterminate forms at x=0. In this case the situation is determined by the limits

(5)p0=p(0)=limx0p(x)=limx0xP(x)

and

(6)q0=q(0)=limx0q(x)=limx0x2Q(x).

If p0=0=q0, then x=0 may be an ordinary point of the differential equation x2y+xp(x)y+q(x)y=0 in (3). Otherwise:

Remark

The most common case in applications, for the differential equation written in the form

(3)y+p(x)xy+q(x)x2y=0,

is that the functions p(x) and q(x) are polynomials. In this case p0=p(0) and q0=q(0) are simply the constant terms of these polynomials, so there is no need to evaluate the limits in Eqs. (5) and (6).

Example 2

To investigate the nature of the point x=0 for the differential equation

x4y+(x2sinx)y+(1cosx)y=0,

we first write it in the form in (3):

y+(sinx)/xxy+(1cosx)/x2x2y=0.

Then l’Hôpital’s rule gives the values

p0=limx0sinxx=limx0cosx1=1

and

q0=limx01cosxx2=limx0sinx2x=12

for the limits in (5) and (6). Since they are not both zero, we see that x=0 is not an ordinary point. But both limits are finite, so the singular point x=0 is regular. Alternatively, we could write

p(x)=sinxx=1x(xx33!+x55!)=1x23!+x45!

and

q(x)=1cosxx2=1x2[1(1x22!+x44!x66!+)]=12!x24!+x46!.

These (convergent) power series show explicitly that p(x) and q(x) are analytic and moreover that p0=p(0)=1 and q0=q(0)=12, thereby verifying directly that x=0 is a regular singular point.

The Method of Frobenius

We now approach the task of actually finding solutions of a second-order linear differential equation near the regular singular point x=0. The simplest such equation is the constant-coefficient equidimensional equation

(7)x2y+p0xy+q0y=0

to which Eq. (3) reduces when p(x)p0 and q(x)q0 are constants. In this case we can verify by direct substitution that the simple power function y(x)=xr is a solution of Eq. (7) if and only if r is a root of the quadratic equation

(8)r(r1)+p0r+q0=0.

In the general case, in which p(x) and q(x) are power series rather than constants, it is a reasonable conjecture that our differential equation might have a solution of the form

(9)y(x)=xrn=0cnxn=n=0cnxn+r=c0xr+c1xr+1+c2xr+2+

—the product of xr and a power series. This turns out to be a very fruitful conjecture; according to Theorem 1 (soon to be stated formally), every equation of the form in (1) having x=0 as a regular singular point does, indeed, have at least one such solution. This fact is the basis for the method of Frobenius, named for the German mathematician Georg Frobenius (1848–1917), who discovered the method in the 1870s.

An infinite series of the form in (9) is called a Frobenius series. Note that a Frobenius series is generally not a power series. For instance, with r=12 the series in (9) takes the form

y=c0x1/2+c1x1/2+c2x3/2+c3x5/2+;

it is not a series in integral powers of x.

To investigate the possible existence of Frobenius series solutions, we begin with the equation

(10)x2y+xp(x)y+q(x)y=0,

obtained by multiplying the equation in (3) by x2. If x=0 is a regular singular point, then p(x) and q(x) are analytic at x=0, so

(11)p(x)=p0+p1x+p2x2+p3x3+,q(x)=q0+q1x+q2x2+q3x3+.

Suppose that Eq. (10) has the Frobenius series solution

(12)y=n=0cnxn+r.

We may (and always do) assume that c00 because the series must have a first nonzero term. Termwise differentiation in Eq. (12) leads to

(13)y=n=0cn(n+r)xn+r1

and

(14)y=n=0cn(n+r)(n+r1)xn+r2.

Substitution of the series in Eqs. (11) through (14) in Eq. (10) now yields

(15)[r(r1)c0xr+(r+1)rc1xr+1+]+[p0x+p1x2+][rc0xr1+(r+1)c1xr+]+[q0+q1x+][c0xr+c1xr+1+]=0.

Upon multiplying initial terms of the two products on the left-hand side here and then collecting coefficients of xr, we see that the lowest-degree term in Eq. (15) is c0[r(r1)+p0r+q0]xr. If Eq. (15) is to be satisfied identically, then the coefficient of this term (as well as those of the higher-degree terms) must vanish. But we are assuming that c00, so it follows that r must satisfy the quadratic equation

(16)r(r1)+p0r+q0=0

of precisely the same form as that obtained with the equidimensional equation in (7). Equation (16) is called the indicial equation of the differential equation in (10), and its two roots (possibly equal) are the exponents of the differential equation (at the regular singular point x=0).

Our derivation of Eq. (16) shows that if the Frobenius series y=xrcnxn is to be a solution of the differential equation in (10), then the exponent r must be one of the roots r1 and r2 of the indicial equation in (16). If r1r2, it follows that there are two possible Frobenius series solutions, whereas if r1=r2 there is only one possible Frobenius series solution; the second solution cannot be a Frobenius series. The exponents r1 and r2 in the possible Frobenius series solutions are determined (using the indicial equation) by the values p0=p(0) and q0=q(0) that we have discussed. In practice, particularly when the coefficients in the differential equation in the original form in (1) are polynomials, the simplest way of finding p0 and q0 is often to write the equation in the form

(17)y+p0+p1x+p2x2+xy+q0+q1x+q2x2+x2y=0.

Then inspection of the series that appear in the two numerators reveals the constants p0 and q0.

Example 3

Find the exponents in the possible Frobenius series solutions of the equation

2x2(1+x)y+3x(1+x)3y(1x2)y=0.

Solution

We divide each term by 2x2(1+x) to recast the differential equation in the form

y+32(1+2x+x2)xy+12(1x)x2y=0,

and thus see that p0=32 and q0=12. Hence the indicial equation is

r(r1)+32r12=r2+12r12=(r+1)(r12)=0,

with roots r1=12 and r2=1. The two possible Frobenius series solutions are then of the forms

y1(x)=x1/2n=0anxnandy2(x)=x1n=0bnxn.

Frobenius Series Solutions

Once the exponents r1 and r2 are known, the coefficients in a Frobenius series solution are determined by substitution of the series in Eqs. (12) through (14) in the differential equation, essentially the same method as was used to determine coefficients in power series solutions in Section 11.2. If the exponents r1 and r2 are complex conjugates, then there always exist two linearly independent Frobenius series solutions. We will restrict our attention here to the case in which r1 and r2 are both real. We also will seek solutions only for x>0. Once such a solution has been found, we need only replace xr1 with |x|r1 to obtain a solution for x<0. The following theorem is proved in Chapter 4 of Coddington’s An Introduction to Ordinary Differential Equations.

We have already seen that if r1=r2, then there can exist only one Frobenius series solution. It turns out that, if r1r2 is a positive integer, then there may or may not exist a second Frobenius series solution of the form in Eq. (19) corresponding to the smaller root r2. Examples 4 through 6 illustrate the process of determining the coefficients in those Frobenius series solutions that are guaranteed by Theorem 1.

Example 4

Find the Frobenius series solutions of

(20)2x2y+3xy(x2+1)y=0.

Solution

First we divide each term by 2x2 to put the equation in the form in (17):

(21)y+32xy+1212x2x2y=0.

We now see that x=0 is a regular singular point, and that p0=32 and q0=12. Because p(x)32 and q(x)=1212x2 are polynomials, the Frobenius series we obtain will converge for all x>0. The indicial equation is

r(r1)+32r12=(r12)(r+1)=0,

so the exponents are r1=12 and r2=1. They do not differ by an integer, so Theorem 1 guarantees the existence of two linearly independent Frobenius series solutions. Rather than separately substituting

y1=x1/2n=0anxnandy2=x1n=0bnxn

in Eq. (20), it is more efficient to begin by substituting y=xrcnxn. We will then get a recurrence relation that depends on r. With the value r1=12 it becomes a recurrence relation for the series for y1, whereas with r2=1 it becomes a recurrence relation for the series for y2.

When we substitute

y=n=0cnxn+r,y=n=0(n+r)cnxn+r1,

and

y=n=0(n+r)(n+r1)cnxn+r2

in Eq. (20)—the original differential equation, rather than Eq. (21)—we get

(22)2n=0(n+r)(n+r1)cnxn+r+3n=0(n+r)cnxn+rn=0cnxn+r+2n=0cnxn+r=0.

At this stage there are several ways to proceed. A good standard practice is to shift indices so that each exponent will be the same as the smallest one present. In this example, we shift the index of summation in the third sum by 2 to reduce its exponent from n+r+2 to n+r. This gives

(23)2n=0(n+r)(n+r1)cnxn+r+3n=0(n+r)cnxn+rn=2cn2xn+rn=0cnxn+r=0.

The common range of summation is n2, so we must treat n=0 and n=1 separately. Following our standard practice, the terms corresponding to n=0 will always give the indicial equation

[2r(r1)+3r1]c0=2(r2+12r12)c0=0.

The terms corresponding to n=1 yield

[2(r+1)r+3(r+1)1]c1=(2r2+5r+2)c1=0.

Because the coefficient 2r2+5r+2 of c1 is nonzero whether r=12 or r=1, it follows that

(24)c1=0

in either case.

The coefficient of xn+r in Eq. (23) is

2(n+r)(n+r1)cn+3(n+r)cncn2cn=0.

We solve for cn and simplify to obtain the recurrence relation

(25)cn=cn22(n+r)n+(n+r)1for n2.

Case 1: r1=12. We now write an in place of cn and substitute r=12 in Eq. (25). This gives the recurrence relation

(26)an=an22n2+3nfor n2.

With this formula we can determine the coefficients in the first Frobenius solution y1. In view of Eq. (24) we see that an=0 whenever n is odd. With n=2, 4, and 6 in Eq. (26), we get

a2=a014,a4=a244=a0616,anda6=a490=a055,440.

Hence the first Frobenius solution is

y1(x)=a0x1/2(1+x214+x4616+x655,440+).

Case 2: r2=1. We now write bn in place of cn and substitute r=1 in Eq. (25). This gives the recurrence relation

(27)bn=bn22b23nfor n2.

Again, Eq. (24) implies that bn=0 for n odd. With n=2, 4, and 6 in (27), we get

b2=b02,b4=b220=b040,andb6=b454=b02160.

Hence the second Frobenius solution is

y2(x)=b0x1(1+x22+x440+x62160+).

Example 5

Find a Frobenius solution of Bessel’s equation of order zero,

(28)x2y+xy+x2y=0.

Solution

In the form of (17), Eq. (28) becomes

y+1xy+x2x2y=0.

Hence x=0 is a regular singular point with p(x)1 and q(x)=x2, so our series will converge for all x>0. Because p0=1 and q0=0, the indicial equation is

r(r1)+r=r2=0.

Thus we obtain only the single exponent r=0, and so there is only one Frobenius series solution

y(x)=x0n=0cnxn

of Eq. (28); it is in fact a power series.

Thus we substitute y=cnxn in (28); the result is

n=0n(n1)cnxn+n=0ncnxn+n=0cnxn+2=0.

We combine the first two sums and shift the index of summation in the third by 2 to obtain

n=0n2cnxn+n=2cn2xn=0.

The term corresponding to x0 gives 0=0: no information. The term corresponding to x1 gives c1=0, and the term for xn yields the recurrence relation

(29)cn=cn2n2for n2.

Because c1=0, we see that cn=0 whenever n is odd. Substituting n=2, 4, and 6 in Eq. (29), we get

c2=c022,c4=c242=c02242,andc6=c462=c0224262.

Evidently, the pattern is

c2n=(1)nc02242(2n)2=(1)nc022n(n!)2.

The choice c0=1 gives us one of the most important special functions in mathematics, the Bessel function of order zero of the first kind, denoted by J0(x). Thus

(30)J0(x)=n=0(1)nx2n22n(n!)2=1x24+x464x62304+.

In this example we have not been able to find a second linearly independent solution of Bessel’s equation of order zero. We will derive that solution in Section 11.4; it will not be a Frobenius series.

When r1-r2 Is an Integer

Recall that, if r1r2 is a positive integer, then Theorem 1 guarantees only the existence of the Frobenius series solution corresponding to the larger exponent r1. Example 6 illustrates the fortunate case in which the series method nevertheless yields a second Frobenius series solution. An example in which the second solution is not a Frobenius series will be discussed in Section 11.4.

Example 6

Find the Frobenius series solutions of

(31)xy+2y+xy=0.

Solution

In standard form the equation becomes

y+2xy+x2x2y=0,

so we see that x=0 is a regular singular point with p0=2 and q0=0. The indicial equation

r(r1)+2r=r(r+1)=0

has roots r1=0 and r2=1, which differ by an integer. In this case when r1r2 is an integer, it is better to depart from the standard procedure of Example 4 and begin our work with the smaller exponent. As you will see, the recurrence relation will then tell us whether or not a second Frobenius series solution exists. If it does exist, our computations will simultaneously yield both Frobenius series solutions. If the second solution does not exist, we begin anew with the larger exponent r=r1 to obtain the one Frobenius series solution guaranteed by Theorem 1.

Hence we begin by substituting

y=x1n=0cnxn=n=0cnxn1

in Eq. (31). This gives

n=0(n1)(n2)cnxn2+2n=0(n1)cnxn2+n=0cnxn=0.

We combine the first two sums and shift the index by 2 in the third to obtain

(32)n=0n(n1)cnxn2+n=2cn2xn2=0.

The cases n=0 and n=1 reduce to

0c0=0and0c1=0.

Hence we have two arbitrary constants c0 and c1 and therefore can expect to find a general solution incorporating two linearly independent Frobenius series solutions. If, for n=1, we had obtained an equation such as 0·c1=3, which can be satisfied for no choice of c1, this would have told us that no second Frobenius series solution could exist.

Now knowing that all is well, from (32) we read the recurrence relation

(33)cn=cn2n(n1)for n2.

The first few values of n give

c2=121c0,c3=132c1,c4=143c2=c04!,c5=154c3=c15!,c6=165c4=c06!,c7=176c6=c17!;

evidently the pattern is

c2n=(1)nc0(2n)!,c2n+1=(1)nc1(2n+1)!

for n1. Therefore, a general solution of Eq. (31) is

y(x)=x1n=0cnxn=c0x(1x22!+x44!)+c1x(xx33!+x55!)=c0xn=0(1)nx2n(2n)!+c1xn=0(1)nx2n+1(2n+1)!.

Thus

y(x)=1x(c0cosx+c1sinx).

We have thus found a general solution expressed as a linear combination of the two Frobenius series solutions

(34)y1(x)=cosxxandy2(x)=sinxx.

As indicated in Fig. 11.3.1, one of these Frobenius series solutions is bounded but the other is unbounded near the regular singular point x=0—a common occurrence in the case of exponents differing by an integer.

FIGURE 11.3.1.

The solutions y1(x)=cosxx and y2(x)=sinxx in Example 6.

Summary

When confronted with a linear second-order differential equation

A(x)y+B(x)y+C(x)y=0

with analytic coefficient functions, in order to investigate the possible existence of series solutions we first write the equation in the standard form

y+P(x)y+Q(x)y=0.

If P(x) and Q(x) are both analytic at x=0, then x=0 is an ordinary point, and the equation has two linearly independent power series solutions.

Otherwise, x=0 is a singular point, and we next write the differential equation in the form

y+p(x)xy+q(x)x2y=0.

If p(x) and q(x) are both analytic at x=0, then x=0 is a regular singular point. In this case we find the two exponents r1 and r2 (assumed real, and with r1r2) by solving the indicial equation

r(r1)+p0r+q0=0,

where p0=p(0) and q0=q(0). There always exists a Frobenius series solution y=xr1anxn associated with the larger exponent r1, and if r1r2 is not an integer, the existence of a second Frobenius series solution y2=xr2bnxn is also guaranteed.

11.3 Problems

In Problems 1 through 8, determine whether x=0 is an ordinary point, a regular singular point, or an irregular singular point. If it is a regular singular point, find the exponents of the differential equation at x=0.

  1. xy+(xx3)y+(sin x)y=0

  2. xy+x2y+(ex1)y=0

  3. x2y+(cos x)y+xy=0

  4. 3x3y+2x2y+(1x2)y=0

  5. x(1+x)y+2y+3xy=0

  6. x2(1x2)y+2xy2y=0

  7. x2y+(6sin x)y+6y=0

  8. (6x2+2x3)y+21xy+9(x21)y=0

If x=a0 is a singular point of a second-order linear differential equation, then the substitution t=xa transforms it into a differential equation having t=0 as a singular point. We then attribute to the original equation at x=a the behavior of the new equation at t=0. Classify (as regular or irregular) the singular points of the differential equations in Problems 9 through 16.

  1. (1x)y+xy+x2y=0

  2. (1x)2y+(2x2)y+y=0

  3. (1x2)y2xy+12y=0

  4. (x2)3y+3(x2)2y+x3y=0

  5. (x24)y+(x2)y+(x+2)y=0

  6. (x29)2y+(x2+9)y+(x2+4)y=0

  7. (x2)2y(x24)y+(x+2)y=0

  8. x3(1x)y+(3x+2)y+xy=0

Find two linearly independent Frobenius series solutions (for x>0) of each of the differential equations in Problems 17 through 26.

  1. 4xy+2y+y=0

  2. 2xy+3yy=0

  3. 2xyyy=0

  4. 3xy+2y+2y=0

  5. 2x2y+xy(1+2x2)y=0

  6. 2x2y+xy(32x2)y=0

  7. 6x2y+7xy(x2+2)y=0

  8. 3x2y+2xy+x2y=0

  9. 2xy+(1+x)y+y=0

  10. 2xy+(12x2)y4xy=0

Use the method of Example 6 to find two linearly independent Frobenius series solutions of the differential equations in Problems 27 through 31. Then construct a graph showing their graphs for x>0.

  1. xy+2y+9xy=0

  2. xy+2y4xy=0

  3. 4xy+8y+xy=0

  4. xyy+4x3y=0

  5. 4x2y4xy+(34x2)y=0

In Problems 32 through 34, find the first three nonzero terms of each of two linearly independent Frobenius series solutions.

  1. 2x2y+x(x+1)y(2x+1)y=0

  2. (2x2+5x3)y+(3xx2)y(1+x)y=0

  3. 2x2y+(sin x)y(cos x)y=0

  4. Note that x=0 is an irregular point of the equation

    x2y+(3x1)y+y=0.
    1. Show that y=xrcnxn can satisfy this equation only if r=0.

    2. Substitute y=cnxn to derive the “formal” solution y=n!xn. What is the radius of convergence of this series?

    1. Suppose that A and B are nonzero constants. Show that the equation x2y+Ay+By=0 has at most one solution of the form y=xrcnxn.

    2. Repeat part (a) with the equation x3y+Axy+By=0.

    3. Show that the equation x3y+Ax2y+By=0 has no Frobenius series solution. (Suggestion: In each case substitute y=xrcnxn in the given equation to determine the possible values of r.)

    1. Use the method of Frobenius to derive the solution y1=x of the equation x3yxy+y=0.

    2. Verify by substitution the second solution y2=xe1/x. Does y2 have a Frobenius series representation?

  5. Apply the method of Frobenius to Bessel’s equation of order 12,

    x2y+xy+(x214)y=0,

    to derive its general solution for x>0,

    y(x)=c0cosxx+c1sinxx.

    Figure 11.3.2 shows the graphs of the two indicated solutions.

    FIGURE 11.3.2.

    The solutions y1(x)=cosxx and y2(x)=sinxx in Problem 38.

    1. Show that Bessel’s equation of order 1,

      x2y+xy+(x21)y=0,

      has exponents r1=1 and r2=1 at x=0, and that the Frobenius series corresponding to r1=1 is

      J1(x)=x2n=0(1)nx2nn!(n+1)!22n.
    2. Show that there is no Frobenius solution corresponding to the smaller exponent r2=1;that is, show that it is impossible to determine the coefficients in

      y2(x)=x1n=0cnxn.
  6. Consider the equation x2y+xy+(1x)y=0.

    1. Show that its exponents are ±i, so it has complex-valued Frobenius series solutions

      y+=xin=0pnxnandy=xin=0qnxn

      with p0=q0=1.

    2. Show that the recursion formula is

      cn=cn1n2+2rn.

      Apply this formula with r=i to obtain pn=cn, then with r=i to obtain qn=cn. Conclude that pn and qn are complex conjugates: pn=an+ibn and qn=anibn, where the numbers {an} and {bn} are real.

    3. Deduce from part (b) that the differential equation given in this problem has real-valued solutions of the form

      y1(x)=A(x)cos(ln x)B(x)sin(ln x),y2(x)=A(x)sin(ln x)+B(x)cos(ln x),

      where A(x)=anxn and B(x)=bnxn.

  7. Consider the differential equation

    x(x1)(x+1)2y+2x(x3)(x+1)y2(x1)y=0

    that appeared in an advertisement for a symbolic algebra program in the March 1984 issue of the American Mathematical Monthly.

    1. Show that x=0 is a regular singular point with exponents r1=1 and r2=0.

    2. It follows from Theorem 1 that this differential equation has a power series solution of the form

      y1(x)=x+c2x2+c3x3+.

      Substitute this series (with c1=1) in the differential equation to show that c2=2, c3=3, and

      cn+2=(n2n)cn1+(n25n2)cn(n2+7n+4)cn+1(n+1)(n+2)

      for n2.

    3. Use the recurrence relation in part (b) to prove by induction that cn=(1)n+1n for n1 (!). Hence deduce (using the geometric series) that

      y1(x)=x(1+x)2

      for 0<x<1.

  8. This problem is a brief introduction to Gauss’s hypergeometric equation

    (35)x(1x)y+[γ(α+β+1)x]yαβy=0,

    where α, β, and γ are constants. This famous equation has wide-ranging applications in mathematics and physics.

    1. Show that x=0 is a regular singular point of Eq. (35), with exponents 0 and 1γ.

    2. If γ is not zero or a negative integer, it follows (why?) that Eq. (35) has a power series solution

      y(x)=x0n=0cnxn=n=0cnxn

      with c00. Show that the recurrence relation for this series is

      cn+1=(α+n)(β+n)(γ+n)(1+n)cn

      for n0.

    3. Conclude that with c0=1 the series in part (b) is

      (36)y(x)=1+n=0αnβnn!γnxn

      where αn=α(α+1)(α+2)(α+n1) for n1, and βn and γn are defined similarly.

    4. The series in (36) is known as the hypergeometric series and is commonly denoted by F(α,β,γ,x). Show that

      1. F(1,1,1,x)=11x (the geometric series);

      2. xF(1,1,2,x)=ln (1+x);

      3. xF(12,1,32,x2)=tan1x;

      4. F(k,1,1,x)=(1+x)k (the binomial series).

11.3 Application Automating the Frobenius Series Method

Here we illustrate the use of a computer algebra system such as Maple to apply the method of Frobenius. More complete versions of this application—illustrating the use of Maple, Mathematica, and Matlab—can be downloaded from our Expanded Applications site using the URL indicated in the margin. We consider the differential equation

(1)2x2y+3xy(x2+1)y=0

of Example 4 in this section, where we found the two indicial roots r1=12 and r2=1.

Beginning with the indicial root r1=12, we first write the initial seven terms of a proposed Frobenius series solution:


a := array(0..6):
y := x^(1/2) ∗sum( a[n] ∗x^(n), n = 0..6);

y:=x(a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6)

Then we substitute this series (actually, partial sum) into the left-hand side of Eq. (1).


deq1 := 2∗x^2∗diff(y,x$2) + 3∗x∗diff(y,x) − (x^2 + 1)∗y:

Evidently x3/2 will factor out upon simplification, so we multiply by x3/2 and then collect coefficients of like powers of x.


deq2 := collect( x^(−3/2)∗simplify(deq1), x);

deq2:=x7a6x6a5+(90a6a4)x5+(a3+65a5)x4+(a2+44a4)x3+(a1+27a3)x2+(14a2a0)x+5a1

We see here the equations that the successive coefficients must satisfy. We can select them automatically by defining an array, then filling the elements of this array by equating to zero (in turn)each of the coefficients in the series.


eqs := array(0..5):
for n from 0 to 5 do
   eqs[n] := coeff(deq1,x,n) = 0: od:
coeffEqs := convert(eqs, set);

coeffEqs:={5a1=0, a2+44a4=0, a3+65a5=0,90a6a4=0, 14a2a0=0, a1+27a3=0}

We now have a collection of six linear equations relating the seven coefficients (a0 through a6). Hence we can proceed to solve for the successive coefficients in terms of a0.


succCoeffs := convert([seq(a[n], n=1..6)], set);
ourCoeffs := solve(coeffEqs, succCoeffs);

ourCoeffs:={a1=0, a6=155440a0, a4=1616a0,a2=114a0, a5=0, a3=0}

Thus we get the first particular solution

y1(x)=a0x1/2(1+x214+x4616+x655440+)

found in Example 4. You can now repeat this process, beginning with the indicial root r2=1, to derive similarly the second particular solution.

In the following problems, use this method to derive Frobenius series solutions that can be checked against the given known general solutions.

  1. xyy+4x3y=0,y(x)=Acosx2+Bsinx2

  2. xy2y+9x5y=0,y(x)=Acosx3+Bsinx3

  3. 4xy2y+y=0,y(x)=Acosx+Bsinx

  4. xy+2y+xy=0,y(x)=1x(Acos x+Bsin x)

  5. 4xy+6y+y=0,y(x)=1x(Acosx+Bsinx)

  6. x2y+xy+(4x41)y=0,y(x)=1x(Acosx2+Bsinx2)

  7. xy+3y+4x3y=0,y(x)=1x2(Acosx2+Bsinx2)

  8. x2y+x2y2y=0,y(x)=1x[A(2x)+B(2+x)ex]

    Problems 9 through 11 involve the arctangent series

    tan1x=xx33+x55x77+.
  9. (x+x3)y+(2+4x2)y+2xy=0,y(x)=1x(A+Btan1x)

  10. (2x+2x2)y+(3+5x)y+y=0,y(x)=1x(A+Btan1x)

  11. (x+x5)y+(3+7x4)y+8x3y=0,y(x)=1x2(A+Btan1x2)

11.4 Bessel Functions

We have already seen several cases of Bessel’s equation of order p0,

(1)x2y+xy+(x2p2)y=0.

Its solutions are now called Bessel functions of order p. Such functions first appeared in the 1730s in the work of Daniel Bernoulli and Euler on the oscillations of a vertically suspended chain. The equation itself appears in a 1764 article by Euler on the vibrations of a circular drumhead, and Fourier used Bessel functions in his classical treatise on heat (1822). But their general properties were first studied systematically in an 1824 memoir by the German astronomer and mathematician Friedrich W. Bessel (1784–1846), who was investigating the motion of planets. The standard source of information on Bessel functions is G. N. Watson’s A Treatise on the Theory of Bessel Functions, 2nd ed. (Cambridge: Cambridge University Press, 1944). Its 36 pages of references, which cover only the period up to 1922, give some idea of the vast literature of this subject.

Bessel’s equation in (1) has indicial equation r2p2=0, with roots r=±p. If we substitute y=cmxm+r in Eq. (1), we find in the usual manner that c1=0 and that

(2)[(m+r)2p2]cm+cm2=0

for m2. The verification of Eq. (2) is left to the reader (Problem 6).

The Case r=p > 0

If we use r=p and write am in place of cm, then Eq. (2) yields the recursion formula

(3)am=am2m(2p+m).

Because a1=0, it follows that am=0 for all odd values of m. The first few even coefficients are

a2=a02(2p+2)=a022(p+1),a4=a24(2p+4)=a0242(p+1)(p+2),a6=a46(2p+6)=a02623(p+1)(p+2)(p+3).

The general pattern is

a2m=(1)ma022mm!(p+1)(p+2)(p+m),

so with the larger root r=p we get the solution

(4)y1(x)=a0m=0(1)mx2m+p22mm!(p+1)(p+2)(p+m).

If p=0 this is the only Frobenius series solution; with a0=1 as well, it is the function J0(x) we have seen before.

The Case r = -p < 0

If we use r=p and write bm in place of cm, Eq. (2) takes the form

(5)m(m2p)bm+bm2=0

for m2, whereas b1=0. We see that there is a potential difficulty if it happens that 2p is a positive integer—that is, if p is either a positive integer or an odd positive integral multiple of 12. For then when m=2p, Eq. (5) is simply 0·bm+bm2=0. Thus if bm20, then no value of bm can satisfy this equation.

But if p is an odd positive integral multiple of 12, we can circumvent this difficulty. For suppose that p=k/2, where k is an odd positive integer. Then we need only choose bm=0 for all odd values of m. The crucial step is the kth step,

k(kk)bk+bk2=0;

and this equation will hold because bk=bk2=0.

Hence if p is not a positive integer, we take bm=0 for m odd and define the coefficients of even subscript in terms of b0 by means of the recursion formula

(6)bm=bm2m(m2p),m2.

In comparing (6) with (3), we see that (6) will lead to the same result as that in (4), except with p replaced with p. Thus in this case we obtain the second solution

(7)y2(x)=b0m=0(1)mx2mp22mm!(p+1)(p+2)(p+m).

The series in (4) and (7) converge for all x>0 because x=0 is the only singular point of Bessel’s equation. If p>0, then the leading term in y1 is a0xp, whereas the leading term in y2 is b0xp. Hence y1(0)=0, but y2(x)± as x0, so it is clear that y1 and y2 are linearly independent solutions of Bessel’s equation of order p>0.

The Gamma Function

The formulas in (4) and (7) can be simplified by use of the gamma function Γ(x), which (as in Section 10.1) is defined for x>0 by

(8)Γ(x)=0ettx1 dt.

It is not difficult to show that this improper integral converges for each x>0. The gamma function is a generalization for x>0 of the factorial function n!, which is defined only if n is a nonnegative integer. To see the way in which Γ(x) is a generalization of n!, we note first that

(9)Γ(1)=0et dt=limb[et]0b=1.

Then we integrate by parts with u=tx and dv=et dt:

Γ(x+1)=limb0bettx dt=limb([ettx]0b+0bxettx1 dt)=x(limb0bettx1 dt);

that is,

(10)Γ(x+1)=xΓ(x).

This is the most important property of the gamma function.

If we combine Eqs. (9) and (10), we see that

Γ(2)=1Γ(1)=1!,Γ(3)=2Γ(2)=2!,Γ(4)=3Γ(3)=3!,

and in general that

(11)Γ(n+1)=n!for n0 an integer.

An important special value of the gamma function is

(12)Γ(12)=0ett1/2 dt=20eu2 du=π,

where we have substituted u2 for t in the first integral; the fact that

0eu2 du=π2

is known, but is far from obvious. (See, for instance, Example 5 in Section 13.4 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008.)

Although Γ(x) is defined in (8) only for x>0, we can use the recursion formula in (10) to define Γ(x) whenever x is neither zero nor a negative integer. If 1<x<0, then

Γ(x)=Γ(x+1)x;

the right-hand side is defined because 0<x+1<1. The same formula may then be used to extend the definition of Γ(x) to the open interval (2,1), then to the open interval (3,2), and so on. The graph of the gamma function thus extended is shown in Fig. 11.4.1. The student who would like to pursue this fascinating topic further should consult Artin’s The Gamma Function (New York: Holt, Rinehart and Winston, 1964). In only 39 pages, this is one of the finest expositions in the entire literature of mathematics.

FIGURE 11.4.1.

The graph of the extended gamma function.

Bessel Functions of the First Kind

If we choose a0=1/[2pΓ(p+1)] in (4), where p>0, and note that

Γ(p+m+1)=(p+m)(p+m1)(p+2)(p+1)Γ(p+1)

by repeated application of Eq. (10), we can write the Bessel function of the first kind of order p very concisely with the aid of the gamma function:

(13)Jp(x)=m=0(1)mm!Γ(p+m+1)(x2)2m+p.

Similarly, if p>0 is not an integer, we choose b0=1/[2pΓ(p+1)] in (7) to obtain the linearly independent second solution

(14)Jp(x)=m=0(1)mm!Γ(p+m+1)(x2)2mp

of Bessel’s equation of order p. If p is not an integer, we have the general solution

(15)y(x)=c1Jp(x)+c2Jp(x)

for x>0; xp must be replaced with |x|p in Eqs. (13) through (15) to get the correct solutions for x<0.

If p=n, a nonnegative integer, then Eq. (13) gives

(16)Jn(x)=m=0(1)mm!(m+n)!(x2)2m+n

for the Bessel functions of the first kind of integral order. Thus

(17)J0(x)=m=0(1)mx2m22m(m!)2=1x222+x42242x6224262+

and

(18)J1(x)=m=0(1)m22m+122m+1m!(m+1)!=x212!(x2)3+12!3!(x2)5.

The graphs of J0(x) and J1(x) are shown in Fig. 11.4.2. In a general way they resemble damped cosine and sine oscillations, respectively. Indeed, if you examine the series in (17), you can see part of the reason why J0(x) and cos x might be similar—only minor changes in the denominators in (17) are needed to produce the Taylor series for cos x. As suggested by Fig. 11.4.2, the zeros of the functions J0(x) and J1(x) are interlaced—between any two consecutive zeros of J0(x) there is precisely one zero of J1(x), and vice versa. The first four zeros of J0(x) are approximately 2.4048, 5.5201, 8.6537, and 11.7915. For n large, the nth zero of J0(x) is approximately (n14)π; the nth zero of J1(x) is approximately (n+14)π. Thus the interval between consecutive zeros of either J0(x) or J1(x) is approximately π—another similarity with cos x and sin x. You can see the way the accuracy of these approximations increases with increasing n by rounding the entries in the table in Fig. 11.4.3 to two decimal places.

It turns out that Jp(x) is an elementary function if the order p is half an odd integer. For instance, on substitution of p=12 in Eqs. (13) and (14), respectively, the results can be recognized (Problem 2) as

(19)J1/2(x)=2πxsinxandJ1/2(x)=2πxcosx.

FIGURE 11.4.2.

The graphs of the Bessel functions J0(x) and J1(x).

FIGURE 11.4.3.

Zeros of J0(x) and J1(x).

n nth zero of J0(x) (n14)π nth zero of J1(x) (n+14)π
1 2.4048 2.3562 3.8317 3.9270
2 5.5201 5.4978 7.0156 7.0686
3 8.6537 8.6394 10.1735 10.2102
4 11.7915 11.7810 13.3237 13.3518
5 14.9309 14.9226 16.4706 16.4934

Bessel Functions of the Second Kind

If n is not an integer, then Bessel’s equation of order n has no second Frobenius series solution independent of Jn(x). For Bessel’s equation of order 0, Example 4 in Section 8.4 of Edwards and Penney, Differential Equations and Boundary Value Problems: Computing and Modeling (5th edition, Hoboken, NJ: Pearson, 2014) gives the second solution

(20)Y0(x)=2π[(γ+lnx2)J0(x)+m=1(1)m+1Hm(m!)2(x2)2m],

where

Hm=1+12+13++1m

denotes the mth partial sum of the harmonic series n=11n and

γ=limn(Hnln n)0.57722

is Euler’s constant. The logarithmic term in (20) implies that this Bessel function Y0(x) of the second kind is not a Frobenius series, and it is typical of the case in which a second Frobenius series solution does not exist. It also implies that Y0(x) as x0 (Fig. 11.4.4), so Y0(x) is not continuous at x=0. These properties are shared by the general Bessel function Yn(x) of the second kind (with n a positive integer), which is defined by a complicated generalization of the formula in (20).

A general solution of Bessel’s equation of integral order n is given by

(21)y(x)=c1Jn(x)+c2Yn(x).

But if y(x) is continuous at x=0, the fact that Yn(x) as x0 implies that c2=0. It follows that any continuous solution of Bessel’s equation of integral order n must be a constant multiple of the Bessel function Jn(x) of the first kind. Numerous physical applications of this fact—to heat flow in circular plates or cylinders and to vibrations of circular membranes, for instance—are discussed in Section 10.4 of the reference cited previously.

Figure 11.4.5 illustrates the fact that for n>1 the graphs of Jn(x) and Yn(x) look generally like those of J1(x) and Y1(x). In particular, Jn(0)=0 while Yn(x) as x0+, and both functions undergo damped oscillation as x+.

FIGURE 11.4.4.

The graphs of the Bessel functions Y0(x) and Y1(x) of the second kind.

FIGURE 11.4.5.

The graphs of the Bessel functions J2(x) and Y2(x).

Bessel Function Identities

Bessel functions are analogous to trigonometric functions in that they satisfy a large number of standard identities of frequent utility, especially in the evaluation of integrals involving Bessel functions. Differentiation of

(13)Jp(x)=m=0(1)mm!Γ(p+m+1)(x2)2m+p

in the case that p is a nonnegative integer gives

ddx[xpJp(x)]=ddxm=0(1)mx2m+2p22m+pm!(p+m)!=m=0(1)mx2m+2p122m+p1m!(p+m1)!=xpm=0(1)mx2m+p122m+p1m!(p+m1)!,

and thus we have shown that

(22)ddx[xpJp(x)]=xpJp1(x).

Similarly,

(23)ddx[xpJp(x)]=xpJp+1(x).

If we carry out the differentiations in Eqs. (22) and (23) and then divide the resulting identities by xp and xp, respectively, we obtain (Problem 8) the identities

(24)Jp(x)=Jp1(x)pxJp(x)

and

(25)Jp(x)=pxJp(x)Jp+1(x).

Thus we may express the derivatives of Bessel functions in terms of Bessel functions themselves. Subtraction of Eq. (25) from Eq. (24) gives the recursion formula

(26)Jp+1(x)=2pxJp(x)Jp1(x),

which can be used to express Bessel functions of higher order in terms of Bessel functions of lower orders. In the form

(27)Jp1(x)=2pxJp(x)Jp+1(x),

it can be used to express Bessel functions of large negative order in terms of Bessel functions of numerically smaller negative orders.

The identities in Eqs. (22) through (27) hold wherever they are meaningful—that is, whenever no Bessel functions of negative integral order appear. In particular, they hold for all nonintegral values of p.

Example 1

With p=0, Eq. (22) gives

xJ0(x) dx=xJ1(x)+C.

Similarly, with p=0, Eq. (23) gives

J1(x) dx=J0(x)+C.

Example 2

Using first p=2 and then p=1 in Eq. (26), we get

J3(x)=4xJ2(x)J1(x)=43[2xJ1(x)J0(x)]J1(x),

so that

J3(x)=4xJ0(x)+(8x21)J1(x).

With similar manipulations every Bessel function of positive integral order can be expressed in terms of J0(x) and J1(x).

Example 3

To antidifferentiate xJ2(x), we first note that

x1J2(x)dx=x1J1(x)+C

by Eq. (23) with p=1. We therefore write

xJ2(x)dx=x2[x1J2(x)]dx

and integrate by parts with

u=x2,dv=x1J2(x)dx,du=2x dx,andv=x1J1(x).

This gives

xJ2(x) dx=xJ1(x)+2J1(x) dx=xJ1(x)2J0(x)+C,

with the aid of the second result of Example 1.

Applications of Bessel Functions

The importance of Bessel functions stems not only from the frequent appearance of Bessel’s equation in applications, but also from the fact that the solutions of many other second-order linear differential equations can be expressed in terms of Bessel functions. To see how this comes about, we begin with Bessel’s equation of order p in the form

(28)z2d2wdz2+zdwdz+(z2p2)w=0

and substitute

(29)w=xαy,z=kxβ.

Then a routine (but lengthy) transformation of Eq. (28) yields

x2y+(12α)xy+(α2β2p2+β2k2x2β)y=0;

that is,

(30)x2y+Axy+(B+Cxq)y=0,

where the constants A, B, C, and q are given by

(31)A=12α,B=α2β2p2,C=β2k2,andq=2β.

It is a simple matter to solve the equations in (31) for

(32)α=1A2,β=q2,k=2Cq,andp=(1A)24Bq.

Under the assumption that the square roots in (32) are real, it follows that the general solution of Eq. (30) is

y(x)=xαw(z)=xαw(kxβ),

where

w(z)=c1Jp(z)+c2Jp(z)

(assuming that p is not an integer) is the general solution of the Bessel equation in (28). This establishes the following result.

Example 4

Solve the equation

(34)4x2y+8xy+(x43)y=0.

Solution

To compare Eq. (34) with Eq. (30), we rewrite the former as

x2y+2xy+(34+14x4)y=0

and see that A=2,B=34,C=14, and q=4. Then the equations in (32) give α=12,β=2,k=14, and p=12. Thus the general solution in (33) of Eq. (34) is

y(x)=x1/2[c1J1/2(14x2)+c2J1/2(14x2)].

If we recall from Eq. (19) that

J1/2(z)=2πzsinzandJ1/2(z)=2πzcosz,

we see that a general solution of Eq. (34) can be written in the elementary form

y(x)=x3/2(Acosx24+Bsinx24).

Example 5

Solve the Airy equation

(35)y+9xy=0.

Solution

First we rewrite the given equation in the form

x2y+9x3y=0.

This is the special case of Eq. (30) with A=B=0,C=9, and q=3. It follows from the equations in (32) that α=12,β=32,k=2, and p=13. Thus the general solution of Eq. (35) is

y(x)=x1/2[c1J1/3(2x3/2)+c2J1/3(2x3/2)].

11.4 Problems

  1. Differentiate termwise the series for J0(x) to show directly that J0(x)=J1(x) (another analogy with the cosine and sine functions).

    1. Deduce from Eqs. (10) and (12) that

      Γ(n+12)=135(2n1)2nπ.
    2. Use the result of part (a) to verify the formulas in Eq. (19) for J1/2(x) and J1/2(x).

    1. Suppose that m is a positive integer. Show that

      Γ(m+23)=258(3m1)3mΓ(23).
    2. Conclude from part (a) and Eq. (13) that

      J1/3(x)=(x/2)1/3Γ(23)(1+m=1(1)m3mx2m22mm!25(3m1)).
  2. Apply Eqs. (19), (26), and (27) to show that

    J3/2(x)=2πx3(sinxxcosx)

    and

    J3/2(x)=2πx3(cosx+xsinx).
  3. Express J4(x) in terms of J0(x) and J1(x).

  4. Derive the recursion formula in Eq. (2) for Bessel’s equation.

  5. Verify the identity in (23) by termwise differentiation.

  6. Deduce the identities in Eqs. (24) and (25) from those in Eqs. (22) and (23).

  7. Use the relation Γ(x+1)=xΓ(x) to deduce from Eqs. (13) and (14) that if p is not a negative integer, then

    Jp(x)=(x/2)pΓ(p+1)[1+m=1(1)m(x/2)2mm!(p+1)(p+2)(p+m)].

    This form is more convenient for the computation of Jp(x) because only the single value Γ(p+1) of the gamma function is required.

  8. Use the series of Problem 9 to find y(0)=limx0y(x) if

    y(x)=x2[J5/2(x)+J5/2(x)J1/2(x)+J1/2(x)].

Any integral of the form xmJn(x) dx can be evaluated in terms of Bessel functions and the indefinite integral J0(x) dx. The latter integral cannot be simplified further, but the function 0xJ0(t) dt is tabulated in Table 11.1 of Abramowitz and Stegun. Use the identities in Eqs. (22) and (23) to evaluate the integrals in Problems 11 through 18.

  1. x2J0(x) dx

  2. x3J0(x) dx

  3. x4J0(x) dx

  4. xJ1(x) dx

  5. x2J1(x) dx

  6. x3J1(x) dx

  7. x4J1(x) dx

  8. J2(x) dx

In Problems 19 through 30, express the general solution of the given differential equation in terms of Bessel functions.

  1. x2yxy+(1+x2)y=0

  2. xy+3y+xy=0

  3. xyy+36x3y=0

  4. x2y5xy+(8+x)y=0

  5. 36x2y+60xy+(9x35)y=0

  6. 16x2y+24xy+(1+144x3)y=0

  7. x2y+3xy+(1+x2)y=0

  8. 4x2y12xy+(15+16x)y=0

  9. 16x2y(5144x3)y=0

  10. 2x2y3xy2(14x5)y=0

  11. y+x4y=0

  12. y+4x3y=0

  13. Apply Theorem 1 to show that the general solution of

    xy+2y+xy=0

    is y(x)=x1(Acosx+Bsinx).

  14. Verify that the substitutions in (2) in Bessel’s equation [Eq. (1)] yield Eq. (3).

    1. Show that the substitution

      y=1ududx

      transforms the Riccati equation dy/dx=x2+y2 into u+x2u=0.

    2. Show that the general solution of dy/dx=x2+y2 is

      y(x)=xJ3/4(12x2)cJ3/4(12x2)cJ1/4(12x2)+J1/4(12x2).

      Suggestion: Apply the identities in Eqs. (22) and (23).

    1. Substitute the series of Problem 9 in the result of Problem 33 to show that the solution of the initial value problem

      dydx=x2+y2,y(0)=0

      is

      y(x)=xJ3/4(12x2)J1/4(12x2).
    2. Deduce similarly that the solution of the initial value problem

      dydx=x2+y2,y(0)=1

      is

      y(x)=x2Γ(34)J3/4(12x2)+Γ(14)J3/4(12x2)2Γ(34)J1/4(12x2)Γ(14)J1/4(12x2).

      Some solution curves of the equation dy/dx=x2+y2 are shown in Fig. 11.4.6. The location of the asymptotes where y(x)+ can be found by using Newton’s method to find the zeros of the denominators in the formulas for the solutions as listed here.

      FIGURE 11.4.6.

      Solution curves of dydx=x2+y2.

References for Further Study

The literature of the theory and applications of differential equations is vast. The following list includes a selection of books that might be useful to readers who wish to pursue further the topics introduced in this book.

  1. 1. Abramowitz, M. and I. A. Stegun, Handbook of Mathematical Functions. New York: Dover, 1965. The comprehensive collection of tables to which frequent reference is made in the text.

  2. 2. Birkhoff, G. and G.-C. Rota, Ordinary Differential Equations (4th ed.). New York: John Wiley, 1989. An intermediate-level text that includes a more complete treatment of existence and uniqueness theorems, Sturm-Liouville problems, and eigenfunction expansions.

  3. 3. Braun, M., Differential Equations and Their Applications (3rd ed.). New York: Springer-Verlag, 1983. An introductory text at a slightly higher level than this book; it includes several interesting ''case study'' applications.

  4. 4. Churchill, R. V., Operational Mathematics (3rd ed.). New York: McGraw-Hill, 1972. The standard reference for theory and applications of Laplace transforms, starting at about the same level as Chapter 10 of this book.

  5. 5. Coddington, E. A., An Introduction to Ordinary Differential Equations. Hoboken, NJ: Pearson, 1961. An intermediate-level introduction; Chapters 3 and 4 include proofs of the theorems on power series and Frobenius series solutions stated in Chapter 11 of this book.

  6. 6. Coddington, E. A. and N. Levinson, Theory of Ordinary Differential Equations. New York: McGraw-Hill, 1955. An advanced theoretical text; Chapter 5 discusses solutions near an irregular singular point.

  7. 7. Dormand, J. R., Numerical Methods for Differential Equations. Boca Raton: CRC Press, 1996. More complete coverage of modern computational methods for approximate solution of differential equations.

  8. 8. Hubbard, J. H. and B. H. West, Differential Equations: A Dynamical Systems Approach. New York: Springer-Verlag, 1992 (part I) and 1995 (Higher-Dimensional Systems). Detailed treatment of qualitative phenomena, with a balanced combination of computational and theoretical viewpoints.

  9. 9. Ince, E. L., Ordinary Differential Equations. New York: Dover, 1956. First published in 1926, this is the classic older reference work on the subject.

  10. 10. McLachlan, N. W., Ordinary Non-Linear Differential Equations in Engineering and Physical Sciences. London: Oxford University Press, 1956. A concrete introduction to the effects of nonlinearity in physical systems.

  11. 11. Noble, B. and J.W. Daniel, Applied Linear Algebra (3rd ed.). Hoboken, NJ: Pearson, 1988. An especially concrete introduction to linear algebra, with significant applications included.

  12. 12. Polking, J. C. and D. Arnold, Ordinary Differential Equations Using MATLAB (3rd ed.). Hoboken, NJ: Pearson, 2003. A manual for using Matlab in an elementary differential equations course; based on the Matlab programs dfield and pplane that are used and referenced in this text.

  13. 13. Press, W. H., B. P. Flannery, S. A. Teukolsky, and W. T. Vetterling, Numerical Recipes: The Art of Scientific Computing, 3rd ed. Cambridge: Cambridge University Press, 2007. Chapter 17 discusses modern techniques for the numerical solution of ordinary differential equations. Programs in C++, as well as C, Fortran, Pascal, and other languages can be downloaded from the accompanying web site numerical.recipes.

  14. 14. Simmons, G. F., Differential Equations (2nd ed.). New York: McGraw-Hill, 1991. An introductory text with interesting historical notes and fascinating applications and with the most eloquent preface in any mathematics book currently in print.

  15. 15. Strang, W. G., Linear Algebra and Its Applications. (4th ed.). New York: Brooks Cole, 2006. An introductory treatment of linear algebra with motivating applications. Appendix B contains a concrete derivation of the Jordan normal form for square matrices.